Chapter 10: Vectors

The displacement is 40 km, \(30°\) East of North.

We represent this as a displacement vector \(\overrightarrow{\mathrm{OA}}\) such that \(|\overrightarrow{\mathrm{OA}}| = 40\) km, and the vector \(\overrightarrow{\mathrm{OA}}\) makes an angle of \(30°\) with the North direction, lying in the East-North quadrant.

Note: In general, “\(\alpha°\) South of West” means a vector in the South-West quadrant that makes an angle of \(\alpha°\) with the West direction.

(i) 10 kg — This measures mass. It has magnitude only and no direction, so it is a scalar.

(ii) 2 meters North-West — This has both a magnitude (2 m) and a direction (North-West), so it is a vector.

(iii) \(40°\) — This measures an angle. It is magnitude only, so it is a scalar.

(iv) 40 Watt — This measures power and has no direction, so it is a scalar.

(v) \(10^{-19}\) coulomb — This measures electric charge. It is magnitude only, so it is a scalar.

(vi) \(20\) m/sec\(^2\) — This measures acceleration, which is the rate of change of velocity and has both magnitude and direction, so it is a vector.

(i) Time period — Scalar

(ii) Distance — Scalar

(iii) Force — Vector

(iv) Velocity — Vector

(v) Work done — Scalar

(i) Coinitial vectors: \(\vec{a}\) and \(\vec{d}\) share the same initial point, so they are coinitial vectors.

(ii) Equal vectors: \(\vec{b}\) and \(\vec{d}\) have the same direction and the same magnitude. Therefore, \(\vec{b}\) and \(\vec{d}\) are equal vectors.

(iii) Collinear but not equal: \(\vec{a}\) and \(\vec{c}\) have parallel lines of support, so they are collinear. However, since they point in opposite directions, they are not equal. Hence \(\vec{a}\) and \(\vec{c}\) are collinear but not equal.

(i) True. \(\vec{a}\) and \(-\vec{a}\) lie along the same line, so they are collinear.

(ii) False. For example, \(\vec{a}\) and \(2\vec{a}\) are collinear, but \(|2\vec{a}| = 2|\vec{a}| \neq |\vec{a}|\).

(iii) False. For example, \(|\hat{i}| = |\hat{j}| = 1\), but \(\hat{i}\) lies along the \(x\)-axis (OX) and \(\hat{j}\) lies along the \(y\)-axis (OY); they are not collinear.

(iv) False. The vectors \(\vec{a}\) and \(-\vec{a}\) are collinear and \(|\vec{a}| = |-\vec{a}|\), but \(\vec{a} \neq -\vec{a}\) because their directions are opposite.

Note: Two vectors \(\vec{a}\) and \(\vec{b}\) are equal if and only if (i) \(|\vec{a}| = |\vec{b}|\) and (ii) \(\vec{a}\) and \(\vec{b}\) have the same direction.

For \(\vec{a} = \hat{i}+\hat{j}+\hat{k}\):

\[|\vec{a}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}\]

For \(\vec{b} = 2\hat{i}-7\hat{j}-3\hat{k}\):

\[|\vec{b}| = \sqrt{4+49+9} = \sqrt{62}\]

For \(\vec{c} = \frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}-\frac{1}{\sqrt{3}}\hat{k}\):

\[|\vec{c}| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{-1}{\sqrt{3}}\right)^2} = \sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}} = \sqrt{1} = 1\]

Let \(\vec{a} = \hat{i}+\hat{j}+\hat{k}\) and \(\vec{b} = \hat{i}+\hat{j}-\hat{k}\).

Clearly \(\vec{a} \neq \vec{b}\), since the coefficients of \(\hat{k}\) differ (\(1 \neq -1\)).

However, their magnitudes are equal:

\[|\vec{a}| = \sqrt{1+1+1} = \sqrt{3} \quad \text{and} \quad |\vec{b}| = \sqrt{1+1+1} = \sqrt{3}\]

\(\therefore |\vec{a}| = |\vec{b}|\).

Remark: Infinitely many such pairs of vectors can be constructed this way.

Let \(\vec{a} = \hat{i}+2\hat{j}+3\hat{k}\) and \(\vec{b} = 2(\hat{i}+2\hat{j}+3\hat{k}) = 2\vec{a}\).

Since \(\vec{b} = m\vec{a}\) where \(m = 2 > 0\), vectors \(\vec{a}\) and \(\vec{b}\) point in the same direction.

But \(\vec{b} \neq \vec{a}\), because \(|\vec{b}| = 2|\vec{a}| \neq |\vec{a}|\).

Remark: Infinitely many such pairs can be formed by choosing any positive scalar multiple.

Given: \(2\hat{i}+3\hat{j} = x\hat{i}+y\hat{j}\).

Comparing coefficients of \(\hat{i}\) and \(\hat{j}\) on both sides:

\[x = 2 \quad \text{and} \quad y = 3\]

Let \(\overrightarrow{\mathrm{AB}}\) be the vector with initial point \(\mathrm{A}(2,1)\) and terminal point \(\mathrm{B}(-5,7)\).

Position vector of A: \(\overrightarrow{\mathrm{OA}} = 2\hat{i}+\hat{j}\);   Position vector of B: \(\overrightarrow{\mathrm{OB}} = -5\hat{i}+7\hat{j}\).

\[\overrightarrow{\mathrm{AB}} = \overrightarrow{\mathrm{OB}} – \overrightarrow{\mathrm{OA}} = (-5\hat{i}+7\hat{j}) – (2\hat{i}+\hat{j}) = -7\hat{i}+6\hat{j}\]

The scalar components of \(\overrightarrow{\mathrm{AB}}\) are the coefficients of \(\hat{i}\) and \(\hat{j}\): \(-7\) and \(6\).

The vector components are \(-7\hat{i}\) and \(6\hat{j}\).

Adding the three vectors component-wise:

\[\vec{a}+\vec{b}+\vec{c} = (1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k} = 0\hat{i}-4\hat{j}-\hat{k} = -4\hat{j}-\hat{k}\]

The unit vector in the direction of \(\vec{a} = \hat{i}+\hat{j}+2\hat{k}\) is given by:

\[\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{1+1+4}} = \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}\]

\[\Rightarrow \quad \hat{a} = \frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}\]

Given \(\mathrm{P}(1,2,3)\) and \(\mathrm{Q}(4,5,6)\), with O as the origin:

\(\overrightarrow{\mathrm{OP}} = \hat{i}+2\hat{j}+3\hat{k}\) and \(\overrightarrow{\mathrm{OQ}} = 4\hat{i}+5\hat{j}+6\hat{k}\).

\[\overrightarrow{\mathrm{PQ}} = \overrightarrow{\mathrm{OQ}} – \overrightarrow{\mathrm{OP}} = 3\hat{i}+3\hat{j}+3\hat{k}\]

The unit vector in the direction of \(\overrightarrow{\mathrm{PQ}}\):

\[\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|} = \frac{3\hat{i}+3\hat{j}+3\hat{k}}{\sqrt{9+9+9}} = \frac{3(\hat{i}+\hat{j}+\hat{k})}{3\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\]

Computing the sum:

\[\vec{a}+\vec{b} = (2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k} = \hat{i}+0\hat{j}+\hat{k}\]

Finding its magnitude:

\[|\vec{a}+\vec{b}| = \sqrt{1+0+1} = \sqrt{2}\]

The required unit vector is:

\[\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|} = \frac{\hat{i}+\hat{k}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}\]

Let \(\vec{a} = 5\hat{i}-\hat{j}+2\hat{k}\).

The required vector of magnitude 8 in the direction of \(\vec{a}\) is:

\[8\hat{a} = 8\cdot\frac{\vec{a}}{|\vec{a}|} = \frac{8(5\hat{i}-\hat{j}+2\hat{k})}{\sqrt{25+1+4}} = \frac{8(5\hat{i}-\hat{j}+2\hat{k})}{\sqrt{30}}\]

\[= \frac{40}{\sqrt{30}}\hat{i} – \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}\]

Let \(\vec{a} = 2\hat{i}-3\hat{j}+4\hat{k}\) and \(\vec{b} = -4\hat{i}+6\hat{j}-8\hat{k}\).

Observe that:

\[\vec{b} = -2(2\hat{i}-3\hat{j}+4\hat{k}) = -2\vec{a}\]

Since \(\vec{b} = m\vec{a}\) where \(m = -2 < 0\), the vectors \(\vec{a}\) and \(\vec{b}\) are collinear (unlike, since \(m < 0\)).

Let \(\vec{a} = \hat{i}+2\hat{j}+3\hat{k}\). The unit vector is:

\[\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{1+4+9}} = \frac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}} = \frac{1}{\sqrt{14}}\hat{i}+\frac{2}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k}\]

The direction cosines are the coefficients of \(\hat{i}, \hat{j}, \hat{k}\) in \(\hat{a}\):

\[\frac{1}{\sqrt{14}},\quad \frac{2}{\sqrt{14}},\quad \frac{3}{\sqrt{14}}\]

Position vector of \(\mathrm{A}(1,2,-3)\) is \(\hat{i}+2\hat{j}-3\hat{k}\); of \(\mathrm{B}(-1,-2,1)\) is \(-\hat{i}-2\hat{j}+\hat{k}\).

\[\overrightarrow{\mathrm{AB}} = (-\hat{i}-2\hat{j}+\hat{k}) – (\hat{i}+2\hat{j}-3\hat{k}) = -2\hat{i}-4\hat{j}+4\hat{k}\]

\[|\overrightarrow{\mathrm{AB}}| = \sqrt{4+16+16} = \sqrt{36} = 6\]

Unit vector along \(\overrightarrow{\mathrm{AB}}\):

\[\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AB}}|} = \frac{-2\hat{i}-4\hat{j}+4\hat{k}}{6} = -\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}\]

The direction cosines are: \(-\dfrac{1}{3},\; -\dfrac{2}{3},\; \dfrac{2}{3}\).

Let \(\vec{a} = \hat{i}+\hat{j}+\hat{k}\). We find the angle \(\theta_1\) between \(\vec{a}\) and OX (represented by \(\hat{i}\)):

\[\cos\theta_1 = \frac{\vec{a}\cdot\hat{i}}{|\vec{a}||\hat{i}|} = \frac{1(1)+1(0)+1(0)}{\sqrt{3}\cdot 1} = \frac{1}{\sqrt{3}} \Rightarrow \theta_1 = \cos^{-1}\frac{1}{\sqrt{3}}\]

By the same calculation, the angle \(\theta_2\) between \(\vec{a}\) and OY (\(\hat{j}\)) is also \(\cos^{-1}\dfrac{1}{\sqrt{3}}\), and the angle \(\theta_3\) between \(\vec{a}\) and OZ (\(\hat{k}\)) is also \(\cos^{-1}\dfrac{1}{\sqrt{3}}\).

\[\therefore \theta_1 = \theta_2 = \theta_3\]

Hence, \(\vec{a} = \hat{i}+\hat{j}+\hat{k}\) is equally inclined to OX, OY and OZ.

Let P.V. of P be \(\vec{a} = \hat{i}+2\hat{j}-\hat{k}\) and P.V. of Q be \(\vec{b} = -\hat{i}+\hat{j}+\hat{k}\).

(i) Internal division in ratio \(2:1\):

P.V. of R \(= \dfrac{m\vec{b}+n\vec{a}}{m+n} = \dfrac{2(-\hat{i}+\hat{j}+\hat{k})+(\hat{i}+2\hat{j}-\hat{k})}{3}\)

\[= \frac{-2\hat{i}+2\hat{j}+2\hat{k}+\hat{i}+2\hat{j}-\hat{k}}{3} = \frac{-\hat{i}+4\hat{j}+\hat{k}}{3} = -\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}+\frac{1}{3}\hat{k}\]

(ii) External division in ratio \(2:1\):

P.V. of R \(= \dfrac{m\vec{b}-n\vec{a}}{m-n} = \dfrac{2(-\hat{i}+\hat{j}+\hat{k})-(\hat{i}+2\hat{j}-\hat{k})}{2-1}\)

\[= -2\hat{i}+2\hat{j}+2\hat{k}-\hat{i}-2\hat{j}+\hat{k} = -3\hat{i}+\hat{k}\]

Remark: Had R divided PQ externally in ratio \(1:2\), it would lie to the left of P with \(\dfrac{PR}{QR} = \dfrac{1}{2}\).

P.V. of \(\mathrm{P}(2,3,4)\): \(\vec{a} = 2\hat{i}+3\hat{j}+4\hat{k}\);   P.V. of \(\mathrm{Q}(4,1,-2)\): \(\vec{b} = 4\hat{i}+\hat{j}-2\hat{k}\).

By the section formula (mid-point), P.V. of midpoint R of PQ is:

\[\frac{\vec{a}+\vec{b}}{2} = \frac{(2\hat{i}+3\hat{j}+4\hat{k})+(4\hat{i}+\hat{j}-2\hat{k})}{2} = \frac{6\hat{i}+4\hat{j}+2\hat{k}}{2} = 3\hat{i}+2\hat{j}+\hat{k}\]

Step I – Finding the side vectors:

\[\overrightarrow{\mathrm{AB}} = \vec{b}-\vec{a} = (2-3)\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k} = -\hat{i}+3\hat{j}+5\hat{k} \quad\cdots(i)\]

\[\overrightarrow{\mathrm{BC}} = \vec{c}-\vec{b} = -\hat{i}-2\hat{j}-6\hat{k} \quad\cdots(ii)\]

\[\overrightarrow{\mathrm{AC}} = \vec{c}-\vec{a} = -2\hat{i}+\hat{j}-\hat{k} \quad\cdots(iii)\]

Since \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}}\), the Triangle Law confirms A, B, C are vertices of a triangle.

Step II – Checking for right angle:

\[\mathrm{AB} = \sqrt{1+9+25} = \sqrt{35}, \quad \mathrm{BC} = \sqrt{1+4+36} = \sqrt{41}, \quad \mathrm{AC} = \sqrt{4+1+1} = \sqrt{6}\]

Checking Pythagoras on the longest side BC:

\[\mathrm{BC}^2 = 41 = 35+6 = \mathrm{AB}^2+\mathrm{AC}^2 \checkmark\]

\(\therefore\) A, B, C are the vertices of a right-angled triangle.

Option (C) is NOT true.

By the Triangle Law of addition: \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}}\), and so \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}} = \overrightarrow{0}\).

For option (C): \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{CA}} = \overrightarrow{\mathrm{AC}}-\overrightarrow{\mathrm{CA}} = \overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AC}} = 2\overrightarrow{\mathrm{AC}} \neq \overrightarrow{0}\).

Option (D) is equivalent to option (A), hence also true.

Option (D) is incorrect. Two collinear vectors can have different directions (opposite directions) and also different magnitudes.

Options (A) and (C) are true by the definition of collinear vectors. Option (B) is a special case of (A) when \(\lambda = \pm 1\).

Given: \(|\vec{a}| = \sqrt{3}\), \(|\vec{b}| = 2\), \(\vec{a}\cdot\vec{b} = \sqrt{6}\).

Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\). Applying the dot product formula:

\[\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{\sqrt{6}}{\sqrt{3}\cdot 2} = \frac{\sqrt{6}}{2\sqrt{3}} = \sqrt{\frac{6}{12}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \cos\frac{\pi}{4}\]

\[\therefore \theta = \frac{\pi}{4}\]

Let \(\vec{a} = \hat{i}-2\hat{j}+3\hat{k}\) and \(\vec{b} = 3\hat{i}-2\hat{j}+\hat{k}\).

\[|\vec{a}| = \sqrt{1+4+9} = \sqrt{14}, \quad |\vec{b}| = \sqrt{9+4+1} = \sqrt{14}\]

\[\vec{a}\cdot\vec{b} = 1(3)+(-2)(-2)+3(1) = 3+4+3 = 10\]

\[\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{10}{\sqrt{14}\cdot\sqrt{14}} = \frac{10}{14} = \frac{5}{7}\]

\[\therefore \theta = \cos^{-1}\frac{5}{7}\]

Let \(\vec{a} = \hat{i}-\hat{j}+0\hat{k}\) and \(\vec{b} = \hat{i}+\hat{j}+0\hat{k}\).

Projection of \(\vec{a}\) on \(\vec{b}\) is:

\[\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \frac{(1)(1)+(-1)(1)+(0)(0)}{\sqrt{1+1}} = \frac{1-1}{\sqrt{2}} = \frac{0}{\sqrt{2}} = 0\]

Remark: A projection of zero confirms that \(\vec{a}\) is perpendicular to \(\vec{b}\).

Let \(\vec{a} = \hat{i}+3\hat{j}+7\hat{k}\) and \(\vec{b} = 7\hat{i}-\hat{j}+8\hat{k}\).

Projection of \(\vec{a}\) on \(\vec{b}\):

\[\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \frac{1(7)+3(-1)+7(8)}{\sqrt{49+1+64}} = \frac{7-3+56}{\sqrt{114}} = \frac{60}{\sqrt{114}}\]

Let \(\vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})\), \(\vec{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})\), \(\vec{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})\).

Unit vector check:

\[|\vec{a}| = \sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}} = \sqrt{\frac{49}{49}} = 1\]

\[|\vec{b}| = \sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}} = \sqrt{\frac{49}{49}} = 1\]

\[|\vec{c}| = \sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}} = \sqrt{\frac{49}{49}} = 1\]

All three are unit vectors.   Mutual perpendicularity check:

\[\vec{a}\cdot\vec{b} = \frac{2\cdot3+3\cdot(-6)+6\cdot2}{49} = \frac{6-18+12}{49} = 0 \Rightarrow \vec{a}\perp\vec{b}\]

\[\vec{b}\cdot\vec{c} = \frac{3\cdot6+(-6)\cdot2+2\cdot(-3)}{49} = \frac{18-12-6}{49} = 0 \Rightarrow \vec{b}\perp\vec{c}\]

\[\vec{a}\cdot\vec{c} = \frac{2\cdot6+3\cdot2+6\cdot(-3)}{49} = \frac{12+6-18}{49} = 0 \Rightarrow \vec{a}\perp\vec{c}\]

Hence, \(\vec{a},\vec{b},\vec{c}\) are mutually perpendicular unit vectors.

Expanding the dot product:

\[(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2-|\vec{b}|^2 = 8 \quad\cdots(ii)\]

Substituting \(|\vec{a}| = 8|\vec{b}|\) into (ii):

\[64|\vec{b}|^2-|\vec{b}|^2 = 8 \Rightarrow 63|\vec{b}|^2 = 8 \Rightarrow |\vec{b}|^2 = \frac{8}{63}\]

\[\Rightarrow |\vec{b}| = \sqrt{\frac{8}{63}} = \sqrt{\frac{4\times2}{9\times7}} = \frac{2}{3}\sqrt{\frac{2}{7}}\]

And from \(|\vec{a}| = 8|\vec{b}|\):

\[|\vec{a}| = 8\cdot\frac{2}{3}\sqrt{\frac{2}{7}} = \frac{16}{3}\sqrt{\frac{2}{7}}\]

Expanding using the distributive property of the dot product:

\[(3\vec{a}-5\vec{b})\cdot(2\vec{a}+7\vec{b}) = 6\vec{a}\cdot\vec{a}+21\vec{a}\cdot\vec{b}-10\vec{b}\cdot\vec{a}-35\vec{b}\cdot\vec{b}\]

Using \(\vec{a}\cdot\vec{a}=|\vec{a}|^2\), \(\vec{b}\cdot\vec{b}=|\vec{b}|^2\), and \(\vec{b}\cdot\vec{a}=\vec{a}\cdot\vec{b}\):

\[= 6|\vec{a}|^2+21\vec{a}\cdot\vec{b}-10\vec{a}\cdot\vec{b}-35|\vec{b}|^2\]

\[= 6|\vec{a}|^2+11\vec{a}\cdot\vec{b}-35|\vec{b}|^2\]

Given: \(|\vec{a}|=|\vec{b}|\), \(\theta = 60°\), and \(\vec{a}\cdot\vec{b} = \dfrac{1}{2}\).

Using the dot product formula:

\[|\vec{a}||\vec{b}|\cos\theta = \frac{1}{2}\]

Substituting \(|\vec{b}|=|\vec{a}|\) and \(\cos60° = \dfrac{1}{2}\):

\[|\vec{a}|^2 \cdot \frac{1}{2} = \frac{1}{2} \Rightarrow |\vec{a}|^2 = 1 \Rightarrow |\vec{a}| = 1\]

(Since magnitudes are never negative.)

\[\therefore \quad |\vec{a}| = |\vec{b}| = 1\]

Given \(\vec{a}\) is a unit vector, so \(|\vec{a}|=1\).

Expanding the given expression:

\[(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = |\vec{x}|^2+\vec{x}\cdot\vec{a}-\vec{a}\cdot\vec{x}-|\vec{a}|^2 = |\vec{x}|^2-|\vec{a}|^2 = 12\]

Substituting \(|\vec{a}|=1\):

\[|\vec{x}|^2-1 = 12 \Rightarrow |\vec{x}|^2 = 13 \Rightarrow |\vec{x}| = \sqrt{13}\]

First, compute \(\vec{a}+\lambda\vec{b}\):

\[\vec{a}+\lambda\vec{b} = (2-\lambda)\hat{i}+(2+2\lambda)\hat{j}+(3+\lambda)\hat{k}\]

Note that \(\vec{c} = 3\hat{i}+\hat{j}+0\hat{k}\).

For perpendicularity, \((\vec{a}+\lambda\vec{b})\cdot\vec{c}=0\):

\[(2-\lambda)(3)+(2+2\lambda)(1)+(3+\lambda)(0) = 0\]

\[6-3\lambda+2+2\lambda = 0 \Rightarrow 8-\lambda = 0 \Rightarrow \lambda = 8\]

Let \(\vec{c} = |\vec{a}|\vec{b}+|\vec{b}|\vec{a} = l\vec{b}+m\vec{a}\) where \(l=|\vec{a}|\) and \(m=|\vec{b}|\).

Let \(\vec{d} = |\vec{a}|\vec{b}-|\vec{b}|\vec{a} = l\vec{b}-m\vec{a}\).

Computing the dot product:

\[\vec{c}\cdot\vec{d} = (l\vec{b}+m\vec{a})\cdot(l\vec{b}-m\vec{a}) = l^2|\vec{b}|^2-\mathrm{lm}\,\vec{b}\cdot\vec{a}+\mathrm{lm}\,\vec{a}\cdot\vec{b}-m^2|\vec{a}|^2\]

\[= l^2|\vec{b}|^2-m^2|\vec{a}|^2\]

Substituting \(l=|\vec{a}|\) and \(m=|\vec{b}|\):

\[= |\vec{a}|^2|\vec{b}|^2-|\vec{b}|^2|\vec{a}|^2 = 0\]

Since \(\vec{c}\cdot\vec{d}=0\), vectors \(\vec{c}\) and \(\vec{d}\) are perpendicular.

From \(\vec{a}\cdot\vec{a}=0 \Rightarrow |\vec{a}|^2=0 \Rightarrow |\vec{a}|=0\), so \(\vec{a}\) is a zero vector.

Now consider \(\vec{a}\cdot\vec{b}=0 \Rightarrow |\vec{a}||\vec{b}|\cos\theta=0\).

Substituting \(|\vec{a}|=0\): \(0\cdot|\vec{b}|\cos\theta = 0\), which is true for any vector \(\vec{b}\).

\(\therefore\) \(\vec{b}\) can be any vector (no restriction is placed on \(\vec{b}\)).

Since \(\vec{a},\vec{b},\vec{c}\) are unit vectors: \(|\vec{a}|=|\vec{b}|=|\vec{c}|=1\).

Squaring both sides of \(\vec{a}+\vec{b}+\vec{c}=\vec{0}\):

\[|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0\]

Substituting \(|\vec{a}|=|\vec{b}|=|\vec{c}|=1\):

\[1+1+1+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0\]

\[2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = -3\]

\[\therefore \quad \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = -\frac{3}{2}\]

Case I: If \(\vec{a}=\vec{0}\), then \(|\vec{a}|=0\), so \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta = 0\).

Case II: If \(\vec{b}=\vec{0}\), then similarly \(\vec{a}\cdot\vec{b}=0\).

Converse is NOT always true. Counter-example:

Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\). Then \(|\vec{a}|=\sqrt{3}\neq 0\), so \(\vec{a}\neq\vec{0}\).

Let \(\vec{b}=\hat{i}+\hat{j}-2\hat{k}\). Then \(|\vec{b}|=\sqrt{6}\neq 0\), so \(\vec{b}\neq\vec{0}\).

But \(\vec{a}\cdot\vec{b} = 1(1)+1(1)+1(-2) = 1+1-2 = 0\).

So \(\vec{a}\cdot\vec{b}=0\) even though neither \(\vec{a}=\vec{0}\) nor \(\vec{b}=\vec{0}\).

P.V. of A: \(\hat{i}+2\hat{j}+3\hat{k}\); of B: \(-\hat{i}\); of C: \(\hat{j}+2\hat{k}\).

\(\angle ABC\) is the angle between \(\overrightarrow{\mathrm{BA}}\) and \(\overrightarrow{\mathrm{BC}}\).

\[\overrightarrow{\mathrm{BA}} = (\hat{i}+2\hat{j}+3\hat{k})-(-\hat{i}) = 2\hat{i}+2\hat{j}+3\hat{k} \quad\cdots(i)\]

\[\overrightarrow{\mathrm{BC}} = (\hat{j}+2\hat{k})-(-\hat{i}) = \hat{i}+\hat{j}+2\hat{k} \quad\cdots(ii)\]

\[\cos\angle ABC = \frac{\overrightarrow{\mathrm{BA}}\cdot\overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BA}}||\overrightarrow{\mathrm{BC}}|} = \frac{2(1)+2(1)+3(2)}{\sqrt{4+4+9}\cdot\sqrt{1+1+4}} = \frac{10}{\sqrt{17}\cdot\sqrt{6}} = \frac{10}{\sqrt{102}}\]

\[\therefore \quad \angle ABC = \cos^{-1}\frac{10}{\sqrt{102}}\]

P.V.’s: \(\overrightarrow{\mathrm{OA}}=\hat{i}+2\hat{j}+7\hat{k}\), \(\overrightarrow{\mathrm{OB}}=2\hat{i}+6\hat{j}+3\hat{k}\), \(\overrightarrow{\mathrm{OC}}=3\hat{i}+10\hat{j}-\hat{k}\).

\[\overrightarrow{\mathrm{AB}} = \overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} = \hat{i}+4\hat{j}-4\hat{k}\]

\[\overrightarrow{\mathrm{AC}} = \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} = 2\hat{i}+8\hat{j}-8\hat{k} = 2(\hat{i}+4\hat{j}-4\hat{k}) = 2\overrightarrow{\mathrm{AB}}\]

Since \(\overrightarrow{\mathrm{AC}} = 2\overrightarrow{\mathrm{AB}}\), i.e., \(\overrightarrow{\mathrm{AC}} = m\overrightarrow{\mathrm{AB}}\), the vectors \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) are parallel. Since they share the common point A, the points A, B, C are collinear.

Let P.V. of A be \(2\hat{i}-\hat{j}+\hat{k}\), B be \(\hat{i}-3\hat{j}-5\hat{k}\), C be \(3\hat{i}-4\hat{j}-4\hat{k}\).

\[\overrightarrow{\mathrm{AB}} = -\hat{i}-2\hat{j}-6\hat{k} \quad\cdots(i)\]

\[\overrightarrow{\mathrm{BC}} = 2\hat{i}-\hat{j}+\hat{k} \quad\cdots(ii)\]

\[\overrightarrow{\mathrm{AC}} = \hat{i}-3\hat{j}-5\hat{k} \quad\cdots(iii)\]

Since \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}}\), the Triangle Law confirms A, B, C form a triangle.

Checking for right angle at C: \(\overrightarrow{\mathrm{BC}}\cdot\overrightarrow{\mathrm{AC}} = 2(1)+(-1)(-3)+1(-5) = 2+3-5 = 0\).

Since \(\overrightarrow{\mathrm{BC}}\perp\overrightarrow{\mathrm{AC}}\), angle C is \(90°\). Therefore \(\triangle ABC\) is right-angled at C.

Since \(\lambda\vec{a}\) is a unit vector:

\[|\lambda\vec{a}| = 1 \Rightarrow |\lambda||\vec{a}| = 1 \Rightarrow |\lambda|\cdot a = 1 \Rightarrow a = \frac{1}{|\lambda|}\]

\(\therefore\) Option (D) is correct.

Computing the cross product by expanding along the first row of the determinant:

\[\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-7&7\\3&-2&2\end{vmatrix}\]

\[= \hat{i}[(-7)(2)-(7)(-2)]-\hat{j}[(1)(2)-(7)(3)]+\hat{k}[(1)(-2)-(-7)(3)]\]

\[= \hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21) = 0\hat{i}+19\hat{j}+19\hat{k}\]

\[\therefore \quad |\vec{a}\times\vec{b}| = \sqrt{0+361+361} = \sqrt{722} = 19\sqrt{2}\]

Compute the two vectors:

\[\vec{c}=\vec{a}+\vec{b}=4\hat{i}+4\hat{j}+0\hat{k} \quad \text{and} \quad \vec{d}=\vec{a}-\vec{b}=2\hat{i}+0\hat{j}+4\hat{k}\]

Their cross product gives a vector perpendicular to both:

\[\vec{n}=\vec{c}\times\vec{d}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4&4&0\\2&0&4\end{vmatrix} = \hat{i}(16-0)-\hat{j}(16-0)+\hat{k}(0-8) = 16\hat{i}-16\hat{j}-8\hat{k}\]

\[|\vec{n}| = \sqrt{256+256+64} = \sqrt{576} = 24\]

Required unit vector:

\[\hat{n} = \pm\frac{\vec{n}}{|\vec{n}|} = \pm\frac{16\hat{i}-16\hat{j}-8\hat{k}}{24} = \pm\left(\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}\right)\]

Let \(\hat{a}=x\hat{i}+y\hat{j}+z\hat{k}\). Since \(|\hat{a}|=1\):

\[x^2+y^2+z^2=1 \quad\cdots(ii)\]

From the angle with \(\hat{i}\): \(\cos\dfrac{\pi}{3} = x \Rightarrow x = \dfrac{1}{2}\).

From the angle with \(\hat{j}\): \(\cos\dfrac{\pi}{4} = y \Rightarrow y = \dfrac{1}{\sqrt{2}}\).

From the angle with \(\hat{k}\): \(\cos\theta = z\).

Substituting into (ii):

\[\frac{1}{4}+\frac{1}{2}+\cos^2\theta=1 \Rightarrow \cos^2\theta = \frac{1}{4} \Rightarrow \cos\theta = \pm\frac{1}{2}\]

Since \(\theta\) is acute, \(\cos\theta > 0\), so \(\cos\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{3}\).

Hence \(z = \dfrac{1}{2}\), and:

\[\hat{a} = \frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k}\]

Components: \(\dfrac{1}{2},\; \dfrac{1}{\sqrt{2}},\; \dfrac{1}{2}\) and acute angle \(\theta=\dfrac{\pi}{3}\).

Expanding the left-hand side using the distributive property of the cross product:

\[\text{L.H.S.} = (\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = \vec{a}\times\vec{a}+\vec{a}\times\vec{b}-\vec{b}\times\vec{a}-\vec{b}\times\vec{b}\]

Using \(\vec{a}\times\vec{a}=\vec{0}\), \(\vec{b}\times\vec{b}=\vec{0}\), and \(\vec{b}\times\vec{a}=-\vec{a}\times\vec{b}\):

\[= \vec{0}+\vec{a}\times\vec{b}+\vec{a}\times\vec{b}-\vec{0} = 2\vec{a}\times\vec{b} = \text{R.H.S.} \quad \checkmark\]

Expanding the cross product:

\[\hat{i}(6\mu-27\lambda)-\hat{j}(2\mu-27)+\hat{k}(2\lambda-6) = 0\hat{i}+0\hat{j}+0\hat{k}\]

Comparing coefficients of \(\hat{i},\hat{j},\hat{k}\):

\[6\mu-27\lambda=0 \quad\cdots(i), \quad 2\mu-27=0 \quad\cdots(ii), \quad 2\lambda-6=0 \quad\cdots(iii)\]

From (ii): \(\mu=\dfrac{27}{2}\).   From (iii): \(\lambda=3\).

Verification in (i): \(6\left(\dfrac{27}{2}\right)-27(3)=81-81=0\). ✓

\[\therefore \quad \lambda=3 \quad \text{and} \quad \mu=\frac{27}{2}\]

From \(\vec{a}\cdot\vec{b}=0\): either \(\vec{a}=\vec{0}\), or \(\vec{b}=\vec{0}\), or \(\vec{a}\perp\vec{b}\).

From \(\vec{a}\times\vec{b}=\vec{0}\): either \(\vec{a}=\vec{0}\), or \(\vec{b}=\vec{0}\), or \(\vec{a}\) and \(\vec{b}\) are collinear/parallel (\(\theta=0\)).

A pair of vectors cannot be simultaneously perpendicular and parallel (unless one is zero). Therefore:

\[\therefore \quad \vec{a}=\vec{0} \quad \text{or} \quad \vec{b}=\vec{0}\]

First note: \(\vec{b}+\vec{c} = (b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}\).

\[\text{L.H.S.} = \vec{a}\times(\vec{b}+\vec{c}) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1+c_1&b_2+c_2&b_3+c_3\end{vmatrix}\]

By the property of determinants (splitting the third row):

\[= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}+\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\c_1&c_2&c_3\end{vmatrix} = \vec{a}\times\vec{b}+\vec{a}\times\vec{c} = \text{R.H.S.} \quad \checkmark\]

If \(|\vec{a}|=0\) or \(|\vec{b}|=0\), then \(|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta=0\), so \(\vec{a}\times\vec{b}=\vec{0}\). ✓

The converse is NOT always true. Counter-example:

Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), so \(|\vec{a}|=\sqrt{3}\neq0\) (non-zero).

Let \(\vec{b}=2(\hat{i}+\hat{j}+\hat{k})=2\hat{i}+2\hat{j}+2\hat{k}\), so \(|\vec{b}|=2\sqrt{3}\neq0\) (non-zero).

But: \(\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\2&2&2\end{vmatrix}=\vec{0}\) (since R\(_2\) and R\(_3\) are proportional).

So \(\vec{a}\times\vec{b}=\vec{0}\) even though neither vector is zero (they are parallel).

P.V. of A: \(\hat{i}+\hat{j}+2\hat{k}\); of B: \(2\hat{i}+3\hat{j}+5\hat{k}\); of C: \(\hat{i}+5\hat{j}+5\hat{k}\).

\[\overrightarrow{\mathrm{AB}} = \hat{i}+2\hat{j}+3\hat{k}, \quad \overrightarrow{\mathrm{AC}} = 0\hat{i}+4\hat{j}+3\hat{k}\]

\[\overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{AC}} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\0&4&3\end{vmatrix} = \hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0) = -6\hat{i}-3\hat{j}+4\hat{k}\]

\[\text{Area} = \frac{1}{2}|\overrightarrow{\mathrm{AB}}\times\overrightarrow{\mathrm{AC}}| = \frac{1}{2}\sqrt{36+9+16} = \frac{1}{2}\sqrt{61} \text{ sq. units}\]

\[\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&-7&1\end{vmatrix} = \hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2) = 20\hat{i}+5\hat{j}-5\hat{k}\]

Area of parallelogram \(=|\vec{a}\times\vec{b}|\):

\[= \sqrt{400+25+25} = \sqrt{450} = \sqrt{225\times2} = 15\sqrt{2} \text{ sq. units}\]

Note: If the diagonals of a parallelogram are \(\vec{\alpha}\) and \(\vec{\beta}\), its area is \(\dfrac{1}{2}|\vec{\alpha}\times\vec{\beta}|\).

Given \(|\vec{a}\times\vec{b}|=1\):

\[|\vec{a}||\vec{b}|\sin\theta = 1 \Rightarrow 3\cdot\frac{\sqrt{2}}{3}\cdot\sin\theta = 1 \Rightarrow \sqrt{2}\sin\theta = 1 \Rightarrow \sin\theta = \frac{1}{\sqrt{2}} = \sin\frac{\pi}{4}\]

\[\Rightarrow \theta = \frac{\pi}{4}\]

\(\therefore\) Option (B) is correct.

Finding the two adjacent sides:

\[\overrightarrow{\mathrm{AB}} = \left(\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}\right)-\left(-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}\right) = 2\hat{i}\]

\[\mathrm{AB} = |\overrightarrow{\mathrm{AB}}| = 2\]

\[\overrightarrow{\mathrm{AD}} = \left(-\hat{i}-\frac{1}{2}\hat{j}+4\hat{k}\right)-\left(-\hat{i}+\frac{1}{2}\hat{j}+4\hat{k}\right) = -\hat{j}\]

\[\mathrm{AD} = |\overrightarrow{\mathrm{AD}}| = 1\]

Area of rectangle \(= \mathrm{AB}\times\mathrm{AD} = 2\times1 = 2\) sq. units.

\(\therefore\) Option (C) is correct.

Let \(\overrightarrow{\mathrm{OP}}\) be the required unit vector in the XY-plane with \(\angle\mathrm{XOP}=30°\), so \(|\overrightarrow{\mathrm{OP}}|=1\).

By the Triangle Law of vector addition in \(\triangle\mathrm{OMP}\):

\[\overrightarrow{\mathrm{OP}} = \mathrm{OP}\cos30°\;\hat{i}+\mathrm{OP}\sin30°\;\hat{j} = (1)\cos30°\;\hat{i}+(1)\sin30°\;\hat{j}\]

\[\Rightarrow \overrightarrow{\mathrm{OP}} = \frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\]

General result: A unit vector making angle \(\theta\) with the positive \(x\)-axis is \((\cos\theta)\hat{i}+(\sin\theta)\hat{j}\).

P.V. of P: \(x_1\hat{i}+y_1\hat{j}+z_1\hat{k}\); P.V. of Q: \(x_2\hat{i}+y_2\hat{j}+z_2\hat{k}\).

\[\overrightarrow{\mathrm{PQ}} = (x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}\]

Scalar components: \((x_2-x_1),\;(y_2-y_1),\;(z_2-z_1)\).

Magnitude:

\[|\overrightarrow{\mathrm{PQ}}| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\]

Taking the departure point as the origin, the first leg of the journey is:

\[\overrightarrow{\mathrm{OA}} = -4\hat{i}\]

The direction “\(30°\) east of north” makes an angle of \(60°\) with the positive \(x\)-axis. Using the general result from Q1, a unit vector in this direction is \(\cos60°\;\hat{i}+\sin60°\;\hat{j}=\dfrac{1}{2}\hat{i}+\dfrac{\sqrt{3}}{2}\hat{j}\).

The second leg of the journey:

\[\overrightarrow{\mathrm{AB}} = 3\left(\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}\right) = \frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\]

Total displacement from O to B (by the Triangle Law):

\[\overrightarrow{\mathrm{OB}} = \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}} = -4\hat{i}+\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j} = -\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\]

The result is not always true.

Case I – Collinear vectors: If \(\vec{a}=\overrightarrow{\mathrm{AC}}\), \(\vec{b}=\overrightarrow{\mathrm{AB}}\), \(\vec{c}=\overrightarrow{\mathrm{BC}}\) are collinear, then \(|\vec{a}| = \mathrm{AC} = \mathrm{AB}+\mathrm{BC} = |\vec{b}|+|\vec{c}|\). This case is true.

Case II – Triangle: When \(\vec{a},\vec{b},\vec{c}\) form the sides of a triangle, the Triangle Law gives \(\vec{a}=\vec{b}+\vec{c}\), but by the triangle inequality: \(|\vec{a}| < |\vec{b}|+|\vec{c}|\). This case is not true.

\(\therefore\; |\vec{a}|=|\vec{b}|+|\vec{c}|\) holds only when \(\vec{b}\) and \(\vec{c}\) are collinear (and point in the same direction).

Since \(x(\hat{i}+\hat{j}+\hat{k}) = x\hat{i}+x\hat{j}+x\hat{k}\) is a unit vector, its magnitude must equal 1:

\[|x\hat{i}+x\hat{j}+x\hat{k}| = 1 \Rightarrow \sqrt{x^2+x^2+x^2} = 1 \Rightarrow \sqrt{3x^2} = 1\]

Squaring: \(3x^2=1 \Rightarrow x^2=\dfrac{1}{3} \Rightarrow x=\pm\dfrac{1}{\sqrt{3}}\).

The resultant vector is:

\[\vec{c}=\vec{a}+\vec{b}=3\hat{i}+\hat{j}+0\hat{k}, \quad |\vec{c}|=\sqrt{9+1}=\sqrt{10}\]

The required vector of magnitude 5 parallel to \(\vec{c}\):

\[5\hat{c} = 5\cdot\frac{\vec{c}}{|\vec{c}|} = \frac{5(3\hat{i}+\hat{j})}{\sqrt{10}} = \frac{5\sqrt{10}}{10}(3\hat{i}+\hat{j}) = \frac{\sqrt{10}}{2}(3\hat{i}+\hat{j}) = \frac{3\sqrt{10}}{2}\hat{i}+\frac{\sqrt{10}}{2}\hat{j}\]

Let \(\vec{d}=2\vec{a}-\vec{b}+3\vec{c}\):

\[= 2(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})+3(\hat{i}-2\hat{j}+\hat{k})\]

\[= 2\hat{i}+2\hat{j}+2\hat{k}-2\hat{i}+\hat{j}-3\hat{k}+3\hat{i}-6\hat{j}+3\hat{k} = 3\hat{i}-3\hat{j}+2\hat{k}\]

The required unit vector:

\[\hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{9+9+4}} = \frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{22}} = \frac{3}{\sqrt{22}}\hat{i}-\frac{3}{\sqrt{22}}\hat{j}+\frac{2}{\sqrt{22}}\hat{k}\]

\[\overrightarrow{\mathrm{AB}} = 4\hat{i}+2\hat{j}+6\hat{k}, \quad \mathrm{AB}=\sqrt{16+4+36}=2\sqrt{14}\]

\[\overrightarrow{\mathrm{BC}} = 6\hat{i}+3\hat{j}+9\hat{k}, \quad \mathrm{BC}=\sqrt{36+9+81}=3\sqrt{14}\]

\[\overrightarrow{\mathrm{AC}} = 10\hat{i}+5\hat{j}+15\hat{k}, \quad \mathrm{AC}=\sqrt{100+25+225}=5\sqrt{14}\]

Since \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}\) and \(\mathrm{AB}+\mathrm{BC}=2\sqrt{14}+3\sqrt{14}=5\sqrt{14}=\mathrm{AC}\), the points A, B, C are collinear.

Finding the ratio: Let B divide AC in ratio \(\lambda:1\). By the section formula:

\[(5,0,-2) = \frac{\lambda(11,3,7)+(1,-2,-8)}{\lambda+1}\]

Comparing coefficients of \(\hat{i}\): \(5(\lambda+1)=11\lambda+1 \Rightarrow -6\lambda=-4 \Rightarrow \lambda=\dfrac{2}{3}\).

The same value follows from the \(\hat{j}\) and \(\hat{k}\) comparisons.

\[\therefore \quad \text{Ratio} = \lambda:1 = \frac{2}{3}:1 = 2:3\]

P.V. of P is \(2\vec{a}+\vec{b}\) and P.V. of Q is \(\vec{a}-3\vec{b}\). For external division in ratio \(1:2=m:n\):

\[\overrightarrow{\mathrm{OR}} = \frac{m\cdot\text{P.V.(Q)}-n\cdot\text{P.V.(P)}}{m-n} = \frac{1(\vec{a}-3\vec{b})-2(2\vec{a}+\vec{b})}{1-2}\]

\[= \frac{\vec{a}-3\vec{b}-4\vec{a}-2\vec{b}}{-1} = \frac{-3\vec{a}-5\vec{b}}{-1} = 3\vec{a}+5\vec{b}\]

Showing P is the midpoint of RQ:

\[\text{P.V. of midpoint of RQ} = \frac{\text{P.V.(R)}+\text{P.V.(Q)}}{2} = \frac{(3\vec{a}+5\vec{b})+(\vec{a}-3\vec{b})}{2} = \frac{4\vec{a}+2\vec{b}}{2} = 2\vec{a}+\vec{b} = \text{P.V. of P}\]

\(\therefore\) Point P is the midpoint of the line segment RQ.

Let \(\vec{a}=2\hat{i}-4\hat{j}+5\hat{k}\) and \(\vec{b}=\hat{i}-2\hat{j}-3\hat{k}\).

The diagonals of the parallelogram are along \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\):

\[\vec{a}+\vec{b} = 3\hat{i}-6\hat{j}+2\hat{k}, \quad |\vec{a}+\vec{b}|=\sqrt{9+36+4}=7\]

\[\vec{a}-\vec{b} = \hat{i}-2\hat{j}+8\hat{k}, \quad |\vec{a}-\vec{b}|=\sqrt{1+4+64}=\sqrt{69}\]

Unit vectors along the diagonals:

\[\frac{3\hat{i}-6\hat{j}+2\hat{k}}{7} \quad \text{and} \quad \frac{\hat{i}-2\hat{j}+8\hat{k}}{\sqrt{69}}\]

Area of parallelogram:

\[\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-4&5\\1&-2&-3\end{vmatrix}=\hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4)=22\hat{i}+11\hat{j}+0\hat{k}\]

\[|\vec{a}\times\vec{b}|=\sqrt{484+121}=\sqrt{605}=\sqrt{121\times5}=11\sqrt{5} \text{ sq. units}\]

Let the vector make equal angles \(\theta, \theta, \theta\) with OX, OY, OZ, with direction cosines \(l, m, n\).

Since the vector is in the positive octant, \(\theta\) is acute. The unit vector is \(\hat{a}=l\hat{i}+m\hat{j}+n\hat{k}\) with \(l^2+m^2+n^2=1\).

For the angle \(\theta\) between \(\hat{a}\) and \(\hat{i}\): \(\cos\theta = l\).

Similarly, \(\cos\theta = m\) and \(\cos\theta = n\). Substituting into the condition:

\[\cos^2\theta+\cos^2\theta+\cos^2\theta=1 \Rightarrow 3\cos^2\theta=1 \Rightarrow \cos\theta=\frac{1}{\sqrt{3}}\]

(Taking the positive value since \(\theta\) is acute.)

\[\therefore \quad l=m=n=\frac{1}{\sqrt{3}}\]

Since \(\vec{a}\times\vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), the required vector is \(\vec{d}=\lambda(\vec{a}\times\vec{b})\) for some scalar \(\lambda\).

\[\vec{d}=\lambda\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&4&2\\3&-2&7\end{vmatrix}=\lambda[\hat{i}(28+4)-\hat{j}(7-6)+\hat{k}(-2-12)]=\lambda(32\hat{i}-\hat{j}-14\hat{k})\]

Using \(\vec{c}\cdot\vec{d}=15\) with \(\vec{c}=2\hat{i}-\hat{j}+4\hat{k}\):

\[2(32\lambda)+(-1)(-\lambda)+4(-14\lambda)=15 \Rightarrow 64\lambda+\lambda-56\lambda=15 \Rightarrow 9\lambda=15 \Rightarrow \lambda=\frac{5}{3}\]

Substituting \(\lambda=\dfrac{5}{3}\):

\[\vec{d}=\frac{5}{3}(32\hat{i}-\hat{j}-14\hat{k})=\frac{1}{3}(160\hat{i}-5\hat{j}-70\hat{k})\]

Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}=2\hat{i}+4\hat{j}-5\hat{k}\), \(\vec{c}=\lambda\hat{i}+2\hat{j}+3\hat{k}\).

\[\vec{d}=\vec{b}+\vec{c}=(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}\]

The unit vector along \(\vec{d}\):

\[\hat{d}=\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda)^2+36+4}}=\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{\lambda^2+4\lambda+44}}\]

Given \(\vec{a}\cdot\hat{d}=1\):

\[\frac{1(2+\lambda)+1(6)+1(-2)}{\sqrt{\lambda^2+4\lambda+44}}=1\]

\[2+\lambda+6-2=\sqrt{\lambda^2+4\lambda+44} \Rightarrow \lambda+6=\sqrt{\lambda^2+4\lambda+44}\]

Squaring both sides:

\[\lambda^2+12\lambda+36=\lambda^2+4\lambda+44 \Rightarrow 8\lambda=8 \Rightarrow \lambda=1\]

Given \(\vec{a},\vec{b},\vec{c}\) are mutually perpendicular with equal magnitudes:

\[\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{a}\cdot\vec{c}=0 \quad\cdots(i)\]

\[|\vec{a}|=|\vec{b}|=|\vec{c}|=\lambda \text{ (say)} \quad\cdots(ii)\]

Let \(\vec{d}=\vec{a}+\vec{b}+\vec{c}\) make angles \(\theta_1,\theta_2,\theta_3\) with \(\vec{a},\vec{b},\vec{c}\) respectively.

Finding \(|\vec{d}|\) first:

\[|\vec{d}|^2=|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{a}\cdot\vec{c})=3\lambda^2\]

\[\therefore |\vec{d}|=\lambda\sqrt{3}\]

Now for \(\theta_1\):

\[\cos\theta_1=\frac{\vec{d}\cdot\vec{a}}{|\vec{d}||\vec{a}|}=\frac{(\vec{a}+\vec{b}+\vec{c})\cdot\vec{a}}{\lambda\sqrt{3}\cdot\lambda}=\frac{|\vec{a}|^2+0+0}{\lambda^2\sqrt{3}}=\frac{\lambda^2}{\lambda^2\sqrt{3}}=\frac{1}{\sqrt{3}}\]

Similarly, \(\cos\theta_2=\dfrac{1}{\sqrt{3}}\) and \(\cos\theta_3=\dfrac{1}{\sqrt{3}}\), giving:

\[\theta_1=\theta_2=\theta_3=\cos^{-1}\frac{1}{\sqrt{3}}\]

\(\therefore\) The vector \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).

First, expand the left-hand side:

\[(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = \vec{a}\cdot\vec{a}+\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{a}+\vec{b}\cdot\vec{b} = |\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2 \quad\cdots(i)\]

If part (given \(\vec{a}\perp\vec{b}\)):

\[\vec{a}\perp\vec{b} \Rightarrow \vec{a}\cdot\vec{b}=0\]

Substituting into (i): \((\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2\). ✓

Only if part (given \((\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2\)):

Substituting this into (i):

\[|\vec{a}|^2+|\vec{b}|^2=|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2 \Rightarrow 0=2\vec{a}\cdot\vec{b} \Rightarrow \vec{a}\cdot\vec{b}=0\]

Since \(\vec{a}\neq\vec{0}\) and \(\vec{b}\neq\vec{0}\), this means \(\vec{a}\) and \(\vec{b}\) are perpendicular. ✓