Chapter 10: Vector Algebra (Vectors and 3D Geometry)

Draw the north line ON from O. Turn 30° towards the east and draw $\overrightarrow{OP}$ of length 4 cm on the scale 1 cm = 10 km. $\overrightarrow{OP}$ represents the displacement.

$\overrightarrow{OP}$: 4 cm long (1 cm = 10 km), 30° east of north.

A vector needs both magnitude and direction. Only (ii) (has a direction, north-west) and (vi) (acceleration) are vectors.

(i) scalar (ii) vector (iii) scalar (iv) scalar (v) scalar (vi) vector

Force and velocity have direction; work done is a dot product (a scalar).

(i) scalar (ii) scalar (iii) vector (iv) vector (v) scalar

From the figure: $\vec{a}$ and $\vec{d}$ start at the same point; $\vec{b}$ and $\vec{d}$ have the same length and the same direction; $\vec{a}$ and $\vec{c}$ are parallel (opposite sides of the square) but point in opposite directions.

(i) $\vec{a}$ and $\vec{d}$ (ii) $\vec{b}$ and $\vec{d}$ (iii) $\vec{a}$ and $\vec{c}$

(i) They lie along the same line. (ii) $\vec{a}$ and $2\vec{a}$ are collinear but unequal. (iii) Equal length says nothing about direction. (iv) They may point in opposite directions ($\vec{a}$ and $-\vec{a}$).

(i) True (ii) False (iii) False (iv) False

$|\vec{a}|=\sqrt{1+1+1}=\sqrt{3}$; $|\vec{b}|=\sqrt{4+49+9}=\sqrt{62}$; $|\vec{c}|=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=1$.

$\sqrt{3}$, $\sqrt{62}$, 1

$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+3\hat{k}$: both have magnitude $\sqrt{14}$ but different directions. (Infinitely many answers are possible.)

e.g. $\hat{i}+2\hat{j}+3\hat{k}$ and $2\hat{i}+\hat{j}+3\hat{k}$

$\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+2\hat{j}+2\hat{k}=2\vec{a}$: the same direction cosines $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$, different magnitudes.

e.g. $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+2\hat{j}+2\hat{k}$

Equal vectors have equal components.

x = 2, y = 3

$\vec{v}=(-5-2)\hat{i}+(7-1)\hat{j}=-7\hat{i}+6\hat{j}$.

Scalar components −7 and 6; vector components $-7\hat{i}$ and $6\hat{j}$.

$(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}$

$-4\hat{j}-\hat{k}$

$|\vec{a}|=\sqrt{1+1+4}=\sqrt{6}$.

$\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}$

$\overrightarrow{PQ}=3\hat{i}+3\hat{j}+3\hat{k}$, $|\overrightarrow{PQ}|=3\sqrt{3}$.

$\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}$

$\vec{a}+\vec{b}=\hat{i}+\hat{k}$, magnitude $\sqrt{2}$.

$\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$

$|\vec{a}|=\sqrt{25+1+4}=\sqrt{30}$. Required vector $=8\hat{a}=\frac{8}{\sqrt{30}}(5\hat{i}-\hat{j}+2\hat{k})$.

$\frac{40}{\sqrt{30}}\hat{i}-\frac{8}{\sqrt{30}}\hat{j}+\frac{16}{\sqrt{30}}\hat{k}$

$-4\hat{i}+6\hat{j}-8\hat{k}=-2(2\hat{i}-3\hat{j}+4\hat{k})$: one is a scalar multiple of the other.

$\vec{b}=-2\vec{a}$, so they are collinear.

$|\vec{a}|=\sqrt{1+4+9}=\sqrt{14}$.

$\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$

$\overrightarrow{AB}=-2\hat{i}-4\hat{j}+4\hat{k}$, $|\overrightarrow{AB}|=\sqrt{4+16+16}=6$.

$-\frac{1}{3},-\frac{2}{3},\frac{2}{3}$

Magnitude $\sqrt{3}$, so $\cos \alpha =\cos \beta =\cos \gamma =\frac{1}{\sqrt{3}}$. The three angles with the axes are equal.

All direction cosines are $\frac{1}{\sqrt{3}}$, so the vector is equally inclined to the axes.

(i) $\vec{r}=\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2\hat{j}-\hat{k})}{2+1}=\frac{-\hat{i}+4\hat{j}+\hat{k}}{3}$

(ii) $\vec{r}=\frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2\hat{j}-\hat{k})}{2-1}=-3\hat{i}+0\hat{j}+3\hat{k}$

(i) $-\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}+\frac{1}{3}\hat{k}$ (ii) $-3\hat{i}+3\hat{k}$

$\frac{(2\hat{i}+3\hat{j}+4\hat{k})+(4\hat{i}+\hat{j}-2\hat{k})}{2}=\frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}$

$3\hat{i}+2\hat{j}+\hat{k}$

$\overrightarrow{AB}=-\hat{i}+3\hat{j}+5\hat{k}$, $|\overrightarrow{AB}{|}^2=35$.

$\overrightarrow{BC}=-\hat{i}-2\hat{j}-6\hat{k}$, $|\overrightarrow{BC}{|}^2=41$.

$\overrightarrow{CA}=2\hat{i}-\hat{j}+\hat{k}$, $|\overrightarrow{CA}{|}^2=6$.

$|\overrightarrow{AB}{|}^2+|\overrightarrow{CA}{|}^2=35+6=41=|\overrightarrow{BC}{|}^2$.

By Pythagoras the triangle is right-angled (at A).

$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}=-\overrightarrow{CA}$, so (C) gives $-2\overrightarrow{CA}\ne \vec{0}$. (A), (B), (D) are all true.

(C)

Collinear only means $\vec{b}=\lambda \vec{a}$. So (A) is correct. (B) is not necessary (λ need not be ±1), (C) is false (components are proportional), (D) is not necessary (they may be opposite or equal).

(B), (C) and (D)

$\cos \theta =\frac{\sqrt{6}}{\sqrt{3}\cdot 2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$.

$\theta =\frac{\pi }{4}$

$\vec{a}\cdot \vec{b}=3+4+3=10$; $|\vec{a}|=|\vec{b}|=\sqrt{14}$. $\cos \theta =\frac{10}{14}=\frac{5}{7}$.

$\theta ={\cos }^{-1}\left(\frac{5}{7}\right)$

$\frac{(\hat{i}-\hat{j})\cdot (\hat{i}+\hat{j})}{|\hat{i}+\hat{j}|}=\frac{1-1}{\sqrt{2}}$

0

$\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}$

$\frac{60}{\sqrt{114}}$

Each magnitude: $\frac{1}{7}\sqrt{4+9+36}=\frac{1}{7}\sqrt{49}=1$ (similarly for the others, $9+36+4=49$ and $36+4+9=49$).

Dot products (ignoring the factor $\frac{1}{49}$): $\vec{a}\cdot \vec{b}=6-18+12=0$; $\vec{b}\cdot \vec{c}=18-12-6=0$; $\vec{c}\cdot \vec{a}=12+6-18=0$.

All are unit vectors and each pair has dot product 0.

$|\vec{a}{|}^2-|\vec{b}{|}^2=8\Rightarrow 64|\vec{b}{|}^2-|\vec{b}{|}^2=8\Rightarrow |\vec{b}{|}^2=\frac{8}{63}$.

$|\vec{b}|=\frac{2\sqrt{2}}{3\sqrt{7}}$, $|\vec{a}|=8|\vec{b}|$.

$|\vec{a}|=\frac{16\sqrt{2}}{3\sqrt{7}}$, $|\vec{b}|=\frac{2\sqrt{2}}{3\sqrt{7}}$

$6\vec{a}\cdot \vec{a}+21\vec{a}\cdot \vec{b}-10\vec{b}\cdot \vec{a}-35\vec{b}\cdot \vec{b}$

$6|\vec{a}{|}^2+11\vec{a}\cdot \vec{b}-35|\vec{b}{|}^2$

$|\vec{a}{|}^2\cos 60^{\circ }=\frac{1}{2}\Rightarrow |\vec{a}{|}^2=1$.

$|\vec{a}|=|\vec{b}|=1$

$|\vec{x}{|}^2-|\vec{a}{|}^2=12\Rightarrow |\vec{x}{|}^2=13$.

$|\vec{x}|=\sqrt{13}$

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

$(\vec{a}+\lambda \vec{b})\cdot \vec{c}=3(2-\lambda )+(2+2\lambda )=8-\lambda =0$.

λ = 8

$(|\vec{a}|\vec{b}+|\vec{b}|\vec{a})\cdot (|\vec{a}|\vec{b}-|\vec{b}|\vec{a})=|\vec{a}{|}^2|\vec{b}{|}^2-|\vec{a}||\vec{b}|\vec{b}\cdot \vec{a}+|\vec{b}||\vec{a}|\vec{a}\cdot \vec{b}-|\vec{b}{|}^2|\vec{a}{|}^2=0$

The dot product is 0, so they are perpendicular.

$\vec{a}\cdot \vec{a}=|\vec{a}{|}^2=0\Rightarrow \vec{a}=\vec{0}$, and then $\vec{a}\cdot \vec{b}=0$ for every $\vec{b}$.

$\vec{b}$ can be any vector.

$|\vec{a}+\vec{b}+\vec{c}{|}^2=0\Rightarrow |\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0\Rightarrow 3+2S=0$.

$-\frac{3}{2}$

Take $\vec{a}=\hat{i}+\hat{j}$ and $\vec{b}=\hat{i}-\hat{j}$: $\vec{a}\cdot \vec{b}=1-1=0$, but neither is the zero vector (they are perpendicular).

e.g. $\vec{a}=\hat{i}+\hat{j}$, $\vec{b}=\hat{i}-\hat{j}$.

$\overrightarrow{BA}=2\hat{i}+2\hat{j}+3\hat{k}$, $\overrightarrow{BC}=\hat{i}+\hat{j}+2\hat{k}$.

$\cos B=\frac{2+2+6}{\sqrt{17}\sqrt{6}}=\frac{10}{\sqrt{102}}$.

$\angle ABC={\cos }^{-1}\left(\frac{10}{\sqrt{102}}\right)$

$\overrightarrow{AB}=\hat{i}+4\hat{j}-4\hat{k}$ and $\overrightarrow{BC}=\hat{i}+4\hat{j}-4\hat{k}$. So $\overrightarrow{AB}=\overrightarrow{BC}$, they are parallel and share point B. (Also $|AB|+|BC|=2\sqrt{33}=|AC|$.)

A, B, C are collinear.

$\overrightarrow{AB}=-\hat{i}-2\hat{j}-6\hat{k}$, $\overrightarrow{BC}=2\hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{CA}=-\hat{i}+3\hat{j}+5\hat{k}$.

$\overrightarrow{BC}\cdot \overrightarrow{CA}=-2-3+5=0$.

BC ⊥ CA, so the triangle is right-angled at C.

$|\lambda \vec{a}|=|\lambda |a=1\Rightarrow a=\frac{1}{|\lambda |}$.

(D)

$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -7& 7\\ 3& -2& 2\end{array}\right|=(-14+14)\hat{i}-(2-21)\hat{j}+(-2+21)\hat{k}=19\hat{j}+19\hat{k}$$

$19\sqrt{2}$

$\vec{a}+\vec{b}=4\hat{i}+4\hat{j}$, $\vec{a}-\vec{b}=2\hat{i}+4\hat{k}$.

$$(\vec{a}+\vec{b})\times (\vec{a}-\vec{b})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 4& 0\\ 2& 0& 4\end{array}\right|=16\hat{i}-16\hat{j}-8\hat{k}$$

Magnitude $=\sqrt{256+256+64}=24$.

$\pm \left(\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}\right)$

${\cos }^2\frac{\pi }{3}+{\cos }^2\frac{\pi }{4}+{\cos }^2\theta =1\Rightarrow \frac{1}{4}+\frac{1}{2}+{\cos }^2\theta =1\Rightarrow \cos \theta =\frac{1}{2}$ (acute).

$\theta =\frac{\pi }{3}$; $\vec{a}=\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k}$

$=\vec{a}\times \vec{a}+\vec{a}\times \vec{b}-\vec{b}\times \vec{a}-\vec{b}\times \vec{b}=\vec{0}+\vec{a}\times \vec{b}+\vec{a}\times \vec{b}-\vec{0}$ (as $\vec{b}\times \vec{a}=-\vec{a}\times \vec{b}$).

$=2(\vec{a}\times \vec{b})$. Proved.

Cross product zero ⇒ the vectors are parallel: $\frac{2}{1}=\frac{6}{\lambda }=\frac{27}{\mu }$.

λ = 3, μ = 27/2

$|\vec{a}||\vec{b}|\cos \theta =0$ and $|\vec{a}||\vec{b}|\sin \theta =0$. Since $\cos \theta$ and $\sin \theta$ cannot both be 0, $|\vec{a}||\vec{b}|=0$.

Either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$.

$$\vec{a}\times (\vec{b}+\vec{c})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1+c_1& b_2+c_2& b_3+c_3\end{array}\right|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|+\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ c_1& c_2& c_3\end{array}\right|$$

(a determinant whose row is a sum splits into two determinants).

$=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$. Proved.

No. Take $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+2\hat{j}+2\hat{k}$ (parallel, both nonzero): $\vec{a}\times \vec{b}=\vec{0}$.

No; any two nonzero collinear vectors give $\vec{a}\times \vec{b}=\vec{0}$.

$\overrightarrow{AB}=\hat{i}+2\hat{j}+3\hat{k}$, $\overrightarrow{AC}=4\hat{j}+3\hat{k}$.

$$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 0& 4& 3\end{array}\right|=-6\hat{i}-3\hat{j}+4\hat{k},|\cdot |=\sqrt{36+9+16}=\sqrt{61}$$

$\frac{\sqrt{61}}{2}$ sq units

$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 3\\ 2& -7& 1\end{array}\right|=20\hat{i}+5\hat{j}-5\hat{k},|\cdot |=\sqrt{400+25+25}=\sqrt{450}$$

$15\sqrt{2}$ sq units

$3\cdot \frac{\sqrt{2}}{3}\sin \theta =1\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}$.

(B)

$\overrightarrow{AB}=2\hat{i}$, $\overrightarrow{BC}=-\hat{j}$. Area $=|\overrightarrow{AB}\times \overrightarrow{BC}|=|-2\hat{k}|=2$.

(C)

$\cos 30^{\circ }\hat{i}+\sin 30^{\circ }\hat{j}$

$\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$

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

Components $x_2-x_1$, $y_2-y_1$, $z_2-z_1$; magnitude $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

Take $\hat{i}$ east, $\hat{j}$ north. First walk: $-4\hat{i}$. Second: $3\sin 30^{\circ }\hat{i}+3\cos 30^{\circ }\hat{j}=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$.

Total: $\left(-4+\frac{3}{2}\right)\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$ (magnitude $\sqrt{\frac{25}{4}+\frac{27}{4}}=\sqrt{13}$ km).

$-\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$

No. If $\vec{a},\vec{b},\vec{c}$ are the sides of a triangle, then $\vec{a}=\vec{b}+\vec{c}$, but by the triangle inequality $|\vec{a}|<|\vec{b}|+|\vec{c}|$. Equality holds only when $\vec{b},\vec{c}$ have the same direction.

Not true in general.

$|x|\sqrt{3}=1$.

$x=\pm \frac{1}{\sqrt{3}}$

Resultant $=3\hat{i}+\hat{j}$, magnitude $\sqrt{10}$. Required vector $=\frac{5}{\sqrt{10}}(3\hat{i}+\hat{j})$.

$\frac{3\sqrt{10}}{2}\hat{i}+\frac{\sqrt{10}}{2}\hat{j}$

$2\vec{a}-\vec{b}+3\vec{c}=(2-2+3)\hat{i}+(2+1-6)\hat{j}+(2-3+3)\hat{k}=3\hat{i}-3\hat{j}+2\hat{k}$, magnitude $\sqrt{22}$.

$\frac{1}{\sqrt{22}}(3\hat{i}-3\hat{j}+2\hat{k})$

$\overrightarrow{AB}=4\hat{i}+2\hat{j}+6\hat{k}$, $\overrightarrow{BC}=6\hat{i}+3\hat{j}+9\hat{k}=\frac{3}{2}\overrightarrow{AB}$. Parallel with common point B: collinear.

$AB:BC=1:\frac{3}{2}=2:3$.

Collinear; B divides AC in the ratio 2 : 3.

$\vec{r}=\frac{1(\vec{a}-3\vec{b})-2(2\vec{a}+\vec{b})}{1-2}=\frac{-3\vec{a}-5\vec{b}}{-1}=3\vec{a}+5\vec{b}$.

Midpoint of RQ $=\frac{(3\vec{a}+5\vec{b})+(\vec{a}-3\vec{b})}{2}=2\vec{a}+\vec{b}$ = position vector of P.

$\vec{r}=3\vec{a}+5\vec{b}$; P is the midpoint of RQ.

Diagonal $=3\hat{i}-6\hat{j}+2\hat{k}$, magnitude $\sqrt{9+36+4}=7$.

$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -4& 5\\ 1& -2& -3\end{array}\right|=22\hat{i}+11\hat{j}+0\hat{k},|\cdot |=\sqrt{605}=11\sqrt{5}$$

Unit vector $\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$; area $11\sqrt{5}$ sq units

Equal inclination ⇒ $l=m=n$. Since $l^2+m^2+n^2=1$: $3l^2=1\Rightarrow l=\pm \frac{1}{\sqrt{3}}$.

Proved.

$\vec{d}$ is parallel to $\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 4& 2\\ 3& -2& 7\end{array}\right|=32\hat{i}-\hat{j}-14\hat{k}$.

$\vec{d}=\mu (32\hat{i}-\hat{j}-14\hat{k})$; $\vec{c}\cdot \vec{d}=\mu (64+1-56)=9\mu =15\Rightarrow \mu =\frac{5}{3}$.

$\vec{d}=\frac{1}{3}(160\hat{i}-5\hat{j}-70\hat{k})$