Question 13 Marks
Find the projection of $\vec{b}+\vec{c}$ on $\vec{a}$ where $\vec{a}=\hat{i}+2\hat{j}+\hat{k},\text{ }\vec{b}=\hat{i}+3\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}+\hat{k}.$
Answer$\vec{\text{b}}+\vec{\text{c}}=2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$
Projection p = $\frac{(2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}})\cdot{(\hat{\text{i}}+2\hat{\text{j}}}+\hat{\text{k}})}{|\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k|}}}$
= $\frac{2 + 6 + 2}{\sqrt{6}}=\frac{10}{\sqrt{6}}\cdot\text{OR }\frac{5\sqrt{6}} {3} $.
View full question & answer→Question 23 Marks
Find the value of $\lambda$ which makes the vectors $\vec{a},\vec{b},\vec{c}$ coplanar, where $\vec{a}=-4\hat{\text{i}}-6\hat{\text{j}}-2\hat{\text{k}},\text{ }\vec{b}=-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{c}=-8\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}.}$
AnswerFor coplanarity $[\vec{a},\vec{b},\vec{c}]=0$
$\therefore\begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & \lambda \end{vmatrix}=0$
$\Rightarrow$ – 8 (–18 + 8) + (–12 – 2) + $\lambda$ (–16 – 6) = 0
$\Rightarrow$ 80 – 14 – 22 $\lambda$ = 0 $\Rightarrow$ $\lambda$ = 3.
View full question & answer→Question 33 Marks
Using vectors, prove that in a $\Delta$ ABC,
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}$
Where a, b and c are lengths of the sides opposite, respectively, to the angles A, B and C of $\Delta$ ABC.
Answer
Let in $\Delta$ ABC, BC = $\vec{\text{a}},\text{ }\text{ CA}=\vec{\text{b}}\text{ and }\text{ }\vec{\text{AB}}=\vec{\text{c}}$
$\therefore\text{ }\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{\text{o}}$
$\vec{\text{a}}\times\vec{\text{a}}+\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{a}}\times\vec{\text{c}}=\vec{\text{o}}\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{c}}\times\vec{\text{a}}............\text{(i)}$
and $\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}=\vec{\text{o}}\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}..........{\text{(ii)}}$
$\therefore\text{ }\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{c}}\times\vec{\text{a}}\Rightarrow\Bigg[\vec{\text{a}}\times\vec{\text{b}}\Bigg]=\Bigg[\vec{\text{b}}\times\vec{\text{c}}\Bigg]=\Bigg[\vec{\text{c}}\times\vec{\text{a}}\Bigg]$
$\therefore\text{ab sin C = bc sin A = ca sin B }\text{ Or }\frac{\sin\text{C}}{\text{c}}=\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}$
$\therefore\frac{\text{a}}{\text{sin A}}=\frac{\text{b}}{\text{sin B}}=\frac{\text{c}}{\text{sin C}}.$ View full question & answer→Question 43 Marks
Find the angle between the vectors $\vec{\text{a}}+\vec{\text{b}}\text{ and }\vec{\text{a}}-\vec{\text{b}}\text{ if }\text{ }\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\text{ and }\vec{\text{b}}\text{ }=3\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}.$
AnswerLet = $\vec{\text{p}}=\vec{\text{a}}+\vec{\text{b}}=\Big(5\hat{\text{i}}+\hat{\text{k}}\Big),\text{ }\vec{\text{p}}=\vec{\text{a}}+\vec{\text{b}}=\Big(-\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}\Big)$
$\text{Using }\text{ }\cos\theta=\frac{\vec{\text{p}}\cdot\vec{\text{q}}}{|\vec{\text{p}}||\vec{\text{q}}|},\text{we get}$
$\cos\theta=0\text{ }\Rightarrow\text{ }\theta=\frac{\pi}{2}.$
View full question & answer→Question 53 Marks
Find the projection of $\overrightarrow{b} + \overrightarrow{c} $ on $\overrightarrow{a}$ where $\overrightarrow{a} = 2\hat{i} - 2\hat{j} + \hat{k}, \overrightarrow{b} = \hat{i} + 2\hat{j} - 2\hat{k} $ and $\overrightarrow{c} = 2\hat{i} - \hat{j} + 4\hat{k}.$
Answer$\overrightarrow{b} + \overrightarrow{c} = 3\hat{i} + \hat{j}+ 2\hat{k} ;\overrightarrow{a} =2\hat{i} - 2\hat{j}+\hat{k} $$\therefore \text{Projection of} (\overrightarrow{b}+\overrightarrow{c}) \text{on} \overrightarrow{a} = \frac{\bigg(3\hat{i}+\hat{j}+2\hat{k}\bigg).{\bigg(2\hat{i} -2\hat{j}+\hat{k}\bigg)}}{\bigg|2\hat{i}-2\hat{j}+\hat{k}\bigg|} $
$\frac{6 - 2 + 2}{3} = 2$
View full question & answer→Question 63 Marks
Find the value of $\lambda$ $\overrightarrow{a} ,\overrightarrow{b}$ and $\overrightarrow{c}$ coplanar, where $\overrightarrow{a} = 2\hat{i} -\hat{j} + \hat{k}, \overrightarrow{b} = \hat{i} + 2\hat{j} - 3\hat{k} $ and $\overrightarrow{c} = 3\hat{i}- \lambda \hat{j} + 5\hat{k}.$
Answer$\overrightarrow{a} = 2\hat{i} -\hat{j} + \hat{k}, \overrightarrow{b} = \hat{i} + 2\hat{j} - 3\hat{k}, \overrightarrow {c} = 3\hat{i}- \lambda \hat{j} + 5\hat{k}.$ are coplanar if $\bigg[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \bigg] = 0$
$ \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & -\lambda & 5 \end{vmatrix} = 0 $
$\Rightarrow 2 (10 - 3 \lambda) + 1 (5 + 9) + ( -\lambda - 6) =0$
$\Rightarrow 7 \lambda = 28$
OR
$\lambda = 4$
View full question & answer→Question 73 Marks
$\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}, \overrightarrow{b} = 3\hat{i} - \hat{j} + 2\hat{k}, \text{show that}\bigg(\overrightarrow{a} +\overrightarrow{b}\bigg) \text{and} \bigg(\overrightarrow{a} -\overrightarrow{b}\bigg)$ are perpendicular to each other.
Answer$(\overrightarrow{a} +\overrightarrow{b}) = 4\hat{i} + \hat{j} - \hat{k} ,\ (\overrightarrow{a} -\overrightarrow{b}) = - 2\hat{i} + 3\hat{j} -5\hat{k}$$\text{For} (\overrightarrow{a} +\overrightarrow{b})\ and\ (\overrightarrow{a} -\overrightarrow{b}) \text{ to be perpendicular } (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a} -\overrightarrow{b})$
$(\overrightarrow{a} +\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b}) = 4(-2) + 1.3 + (-1)(-5)$
$= - 8 + 3 + 5 = 0$
$\therefore (\overrightarrow{a}+\overrightarrow{b}) \perp (\overrightarrow{a}-\overrightarrow{b})$
View full question & answer→Question 83 Marks
Using vectors prove that the line segment joining the mid-points of non-parallel sides of a trapezium is parallel to the base and is equal to half the sum of the parallel sides.
AnswerLet $\overrightarrow{a}, \overrightarrow{b} \text{and}\overrightarrow{c}$ be the position vectors of A, B and C respectively w.r.t. O. Let D and E be the mid-points of parallel sides OC and AB respectively.
$\therefore \text{Position vector of}$
$\text{D is} \frac{O +\overrightarrow{C}}{2} = \frac{\overrightarrow{c}}{2}$
$CB\parallel OA \Rightarrow \overrightarrow{a}\parallel \overrightarrow{b} - \overrightarrow{c}$
$\Rightarrow \text{E is parallel to base and its length is half the sum of lengths of the parallel sides.}$ View full question & answer→Question 93 Marks
Find the coordinates of the tip of the position vector which is equivalent to $\overrightarrow{\text{AB}}$, where the coordinates of A and B are (-1, 3) and (-2, 1) respectively.
AnswerLet O be the origin. Let P(x, y) be the required point. Then $\vec{\text{P}}$ is the tip of the position vector $\overrightarrow{\text{OP}}$ of the point P. We have,$\overrightarrow{\text{OP}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}$
And, $\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A $=\big(-2\hat{\text{i}}+\hat{\text{j}}\big)-\big(-\hat{\text{i}}+3\hat{\text{j}}\big)$ $=-2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{i}}-3\hat{\text{j}}$ $=-\hat{\text{i}}-2\hat{\text{j}}$ Given that $\overrightarrow{\text{OP}}=\overrightarrow{\text{AB}}$ So, $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}=-\hat{\text{i}}-2\hat{\text{j}}\Leftrightarrow\text{x}=-1,\ \text{y}=-2$ Hence, coordinated of the required point is (-1, -2)
View full question & answer→Question 103 Marks
Write two different vectors having same direction.
AnswerConsider $\vec{P}=\Big(\hat{i}+\hat{j}+\hat{k}\Big)\ \text{and}\ \vec{q}=\Big(2\hat{i}+2\hat{j}+2\hat{k}\Big).$The direction cosines of $\vec{p}$ are given by,
$I=\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}},$ $m=\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}},$ $\text{and}\ n=\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}},$ The direction cosines of $\vec{q}$ are given by,$I=\frac{2}{\sqrt{2^2+2^2+2^2}}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}},$ $m=\frac{2}{\sqrt{2^2+2^2+2^2}}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}},$ $\text{and}\ n=\frac{2}{\sqrt{2^2+2^2+2^2}}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}},$
The direction cosines of $\vec{p}\ \text{and}\ \vec{q}$ are the same. Hence, the tow vectors have the same direction.
View full question & answer→Question 113 Marks
Show that $\big|\vec{a}|\ \vec{b}+\big|\vec{b}\big|\ \vec{a}$ is perpendicular to $\big|\vec{a}|\ \vec{b}-\big|\vec{b}\big|\ \vec{a},$ for any two nonzero vectors $\vec{a}\ \text{and}\ \vec{b}.$
Answer$\text{Let}\ \vec{c}=\big|\vec{a}\big|\ \vec{b}+\big|\vec{b}\big|\ \vec{a}=l\vec{b}+m\vec{a},$ $\text{where}\ l=\big|\vec{a}\big|\ \text{and}\ m=\big|\vec{b}\big|$
$\text{Let}\ \ \ \ \ \ \vec{d}=\big|\vec{a}\big|\ \vec{b}-\big|\vec{b}\big|\ \vec{a}=l\vec{b}-m\vec{a}$
$\text{Now}\ \ \ \ \ \ \big|\vec{c}\big|.\Big|\vec{d}\Big|=\big(l\vec{b}+m\vec{a}\big)\big(l\vec{b}-m\vec{a}\big)$ $=l^2\vec{b}.\vec{b}-lm\ \vec{b}.\vec{a}+lm\vec{a}.\vec{b}-m^2\vec{a}.\vec{a}$
$=l^2\Big|\vec{b}\Big|^2-lm\vec{a}.\vec{b}+lm\vec{a}.\vec{b}-m^2\big|\vec{a}|^2$
$=l^2\Big|\vec{b}\Big|^2-m^2\big|\vec{a}|^2$
$\text{Putting},\ l=\big|\vec{a}\big|\ \text{and}\ m=\Big|\vec{b}\Big|,$
$=\big|\vec{a}\big|^2\big|\vec{b}\big|^2-\big|\vec{b}\big|^2\cdot|\vec{a}|^2$
$\Rightarrow\ \ \ \ \ \ \big|\vec{c}\big|.\Big|\vec{d}\Big|=0$
View full question & answer→Question 123 Marks
Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.
AnswerGiven: A regular octagon of eight sides with center O.
To show: $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}+\overrightarrow{\text{OG}}+\overrightarrow{\text{OH}}=\vec0$.
Proof: We know center of the regular octagon bisects all the diagonals passing through it.
$\overrightarrow{\text{OA}}=-\overrightarrow{\text{OE}},\ \overrightarrow{\text{OB}}=-\overrightarrow{\text{OF}},\ \overrightarrow{\text{OD}}=-\overrightarrow{\text{OH}}$ and $\overrightarrow{\text{OC}}=-\overrightarrow{\text{OG}}.$
$\Rightarrow\overrightarrow{\text{OA}}+\overrightarrow{\text{OE}}=\vec0,\ \overrightarrow{\text{OB}}+\overrightarrow{\text{OF}}=\vec0,\ \overrightarrow{\text{OD}}+\overrightarrow{\text{OH}}=\vec0$ and $\overrightarrow{\text{OC}}+\overrightarrow{\text{OG}}=\vec0.\ \dots(\text{i})$
Now,
$\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}+\overrightarrow{\text{OG}}+\overrightarrow{\text{OH}}$
$=\Big(\overrightarrow{\text{OA}}+\overrightarrow{\text{OE}}\Big)+\Big(\overrightarrow{\text{OB}}+\overrightarrow{\text{OF}}\Big)+\Big(\overrightarrow{\text{OC}}+\overrightarrow{\text{OG}}\Big)+\Big(\overrightarrow{\text{OD}}+\overrightarrow{\text{OH}}\Big)$
$=\vec0+\vec0+\vec0+\vec0$
$=\vec0$
Hence proved.
View full question & answer→Question 133 Marks
$\text{Let}\ \vec{\text{a}}=\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}=3\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}$ and $ \vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$ Find a vector $\vec{\text{d}}$ which is perpendicular to both $\vec{\text{a}}\ \text{and}\ \vec{\text{b}},\ \text{and}\ \vec{\text{c}}\cdot\vec{\text{d}}=15.$
Answer$\text{Let}\ \vec{\text{d}}=\text{d}_{1}\hat{\text{i}}+\text{d}_{2}\hat{\text{j}}+\text{d}_{3}\hat{\text{k}}$ Since $\vec{\text{d}}$ is perpendicular to both $\vec{\text{a}}\ \text{and}\ \vec{\text{b}},$ we have: $\vec{\text{d}}\cdot\vec{\text{a}}=0$ $\Rightarrow \text{d}_{1}+4\text{d}_{2}+2\text{d}_{3}=0\ \ \ ...\text{(i)}$ And, $\vec{\text{d}}\cdot\vec{\text{b}}=0$ $\Rightarrow 3\text{d}_{1}-2\text{d}_{2}+7\text{d}_{3}=0\ \ \ ...\text{(ii)}$ Also, it is given that: $\vec{\text{c}}\cdot\vec{\text{d}}=15$ $\Rightarrow 2\text{d}_{1}-\text{d}_{2}+4\text{d}_{3}=15\ \ \ ...\text{(iii)}$ On solving (i), (ii), and (iii), we get: $\text{d}_{1}=\frac{160}{3},\text{d}_{2}=-\frac{5}{3}\ \text{and}\ \text{d}_{3}=-\frac{70}{3}$ $\therefore\vec{\text{d}}=\frac{160}{3}\hat{\text{i}}-\frac{5}{3}\hat{\text{j}}-\frac{70}{3}\hat{\text{k}}=\frac{1}{3}\Big(160\hat{\text{i}}-5\hat{\text{j}}-70\hat{\text{k}}\Big)$Hence, the required vector is $\frac{1}{3}\Big(160\hat{\text{i}}-5\hat{\text{j}}-70\hat{\text{k}}\Big)$
View full question & answer→Question 143 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ of magnitudes 3 and 4 respectively, write the value of $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
AnswerWe have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$ (By definition of scalar triple product)
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\theta\big)^2+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta\big)^2$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big)^2\sin^2\theta+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big)^2\cos^2\theta$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big)^2(\sin^2\theta+\cos^2\theta)$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big)^2$ $\big(\therefore\sin^2\theta+\cos^2\theta=1\big)$
$=(3\times4)^2$ $\text{(Given}:\big|\vec{\text{a}}\big|=3\text{ and } \big|\vec{\text{b}}\big|=4\big)$ $$
$=144$
View full question & answer→Question 153 Marks
Find the position vector of a point R which divides the line joining the two points P and Q with position vectors $\overrightarrow{\text{OP}}=2\vec{\text{a}}+\vec{\text{b}}\text{ and }\overrightarrow{\text{OQ}}=\vec{\text{a}}-2\vec{\text{b}}$, respectively in the ratio 1 : 2 internally and externally.
AnswerIt is given that P and Q are two points with position vectors $\overrightarrow{\text{OP}}=2\vec{\text{a}}+\vec{\text{b}}\text{ and }\overrightarrow{\text{OQ}}=\vec{\text{a}}-2\vec{\text{b}}$, respectively.
When R divides PQ internally in the ratio 1 : 2, then
Position vector of $\text{R}=\frac{1\times\big(\vec{\text{a}}-2\vec{\text{b}}\big)+2\times\big(2\vec{\text{a}}+\vec{\text{b}}\big)}{1+2}=\frac{5\vec{\text{a}}}{3}$
When R divides PQ externally in the ratio 1 : 2, then
Position vector of $\text{R}=\frac{1\times\big(\vec{\text{a}}-2\vec{\text{b}}\big)-2\times\big(2\vec{\text{a}}+\vec{\text{b}}\big)}{1-2}=\frac{-3\vec{\text{a}}-4\vec{\text{b}}}{-1}=3\vec{\text{a}}+4\vec{\text{b}}$
View full question & answer→Question 163 Marks
Prove that the given vectors are coplanar:
$\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ 2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
AnswerGiven the vectors $\text{P}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big),\ \text{Q}\big(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\big)$ and $\text{R}\big(-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$
We know the three vectors are coplanar if oneof them is expressible as a linear combination of the other two. Let,
$\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}=\text{x}\big(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\big)+\text{y}\big(-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$
$=\hat{\text{i}}(2\text{x}-\text{y})+\hat{\text{j}}(3\text{x}-2\text{y})+\hat{\text{k}}(-\text{x}+2\text{y})$
$\Rightarrow2\text{x}-\text{y}=1,\ 3\text{x}-2\text{y}=1,\ -\text{x}+2\text{y}=1$ [Equating the coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ respectively]
Solving first two of these equation, we get x = 1, y = 1. Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.
View full question & answer→Question 173 Marks
ABCDE is a pentagon, prove that,$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}=3\ \overrightarrow{\text{AC}}$
AnswerIt is given that ABCDE is a pentagon, So
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}$
$=\Big(\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\Big)+\overrightarrow{\text{AE}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\Big(\overrightarrow{\text{AE}}+\overrightarrow{\text{ED}}\Big)+\overrightarrow{\text{AC}}$ $\Big[$Using triangle law in $\triangle\text{ABC},\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\Big(\overrightarrow{\text{AD}}\Big)+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}-\overrightarrow{\text{DA}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}$
$=3\ \overrightarrow{\text{AC}}$
So,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}=3\ \overrightarrow{\text{AC}}$
View full question & answer→Question 183 Marks
Find the vector from the origin O to the centroid of the triangle whose vertices are (1, -1, 2), (2, 1, 3) and (-1, 2, -1).
AnswerGiven the vertices of the triangle (1, -1, 2), (2, 1, 3) and (-1, 2, -1). Then, Position vectors are: $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ $\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ The centroid of a triangle is given by $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}$ So, $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}=\frac{\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}+2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}3{}$ $=\frac{2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}}3=\frac{2}3\hat{\text{i}}+\frac{2}3\hat{\text{j}}+\frac{4}3\hat{\text{k}}$
View full question & answer→Question 193 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{b}}\big|,$ then prove that $\vec{\text{a}}+2\vec{\text{b}}$ is perpendicular to $\vec{\text{a}}.$
AnswerGiven that
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{b}}\big|$
Squaring both sides, we get
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2+2\vec{\text{a}}.\vec{\text{b}}=0\dots(1)$
Now,
$\big(\vec{\text{a}}+2\vec{\text{b}}\big).\vec{\text{a}}$
$\vec{\text{a}}.\vec{\text{a}}+2\vec{\text{b}}.\vec{\text{a}}$
$=|\vec{\text{a}}|^2+2\vec{\text{a}}.\vec{\text{b}}$
$=0$ [Using (1)]
So, $\vec{\text{a}}+2\vec{\text{b}}$ is perpendicular to $\vec{\text{a}}.$
View full question & answer→Question 203 Marks
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors:
$\vec{\text{a}}=11\hat{\text{i}},\vec{\text{b}}=2\hat{\text{j}},\vec{\text{c}}=13\hat{\text{k}}$
AnswerGiven,
$\vec{\text{a}}=11\hat{\text{i}}$
$\vec{\text{b}}=2\hat{\text{j}}$
$\vec{\text{c}}=13\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjacent edges are $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$
is equal to $\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|$ Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}11&0&0\\0&2&0\\0&0&13 \end{vmatrix}$
$=11(26-0)-0(0-0)+0(0-0)$
$=286$
volume of the parallelopiped $=\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|=|286|=286$ cubic units.
View full question & answer→Question 213 Marks
Find a unit vector perpendicular to both the vectors $4\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $-2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}.}$
AnswerA vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{b}}.$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$\vec{\text{c}}(\text{say})=\begin{bmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-1&3\\-2&1&-2 \end{bmatrix}$
$\vec{\text{c}}=\hat{\text{i}}(2-3)-\hat{\text{j}}(-8+6)+\hat{\text{k}}(4-2)$
$\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{c}}$ is a vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\hat{\text{c}}=\frac{\vec{\text{c}}}{|\vec{\text{c}}|}$
$=\frac{-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{(-1)^2+(2)^2+(2)^2}}$
$=\frac{-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{1+4+4}}$
$=\frac{1}{3}\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$
So, unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\frac{1}{3}\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$
View full question & answer→Question 223 Marks
Prove that the given vectors are coplanar:
$2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\ \hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}$ and $3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}$
AnswerGiven the vectors $\text{P}\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big),\ \text{Q}\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big)$ and $\text{R}\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$
We know the three vectors are coplanar if oneof them is expressible as a linear combination of the other two. Let,
$2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}=\text{x}\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big)+\text{y}\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$
$=\hat{\text{i}}(\text{x}+3\text{y})+\hat{\text{j}}(-3\text{x}-4\text{y})+\hat{\text{k}}(-5\text{x}-4\text{y})$
$\Rightarrow\text{x}+3\text{y}=2,\ -3\text{x}-4\text{y}=-1,\ -5\text{x}-4\text{y}=1$ [Equating the coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ respectively]
Solving first two of these equation, we get x = -1, y = 1. Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.
View full question & answer→Question 233 Marks
ABCD is a parallelogram. If the coordinates of A, B, C are (-2, -1), (3, 0) and (1, -2) respectively, find the coordinates of D.
AnswerLet the coordinates of D is (x, y).
Since, ABCD is a parallelogram.
$\therefore$ AB = DC
We have,
$\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$
$\Rightarrow3\hat{\text{i}}-\big(-2\hat{\text{i}}-\hat{\text{j}}\big)=\big(\hat{\text{i}}-2\hat{\text{j}}\big)-\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}\big)$
$\Rightarrow5\hat{\text{i}}+\hat{\text{j}}=\hat{\text{i}}(1-\text{x})+\hat{\text{j}}(-2-\text{y})$
$\Rightarrow1-\text{x}=5\text{ and }1=-2-\text{y}$
$\Rightarrow\text{x}=-4\text{ and }\text{y}=-3$
Hence, the coordinates of D is (-4, -3)
View full question & answer→Question 243 Marks
If $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},$and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}$ are such that $\vec{\text{a}}+\lambda\vec{\text{b}}$ is perpendicular to $\vec{\text{c}},$ then find the value of $\lambda.$
AnswerThe given vectors are $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}$.
Now,
$\vec{\text{a}}+\lambda\vec{\text{b}}=\big(2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)\\=(2-\lambda)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(3+\lambda)\hat{\text{k}}$
If $\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)$ is perpendicular to $\vec{\text{c}},$ then
$\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big).\vec{\text{c}}=0.$
$\Rightarrow\Big[\big(2-\lambda\big)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(3+\lambda)\hat{\text{k}}\Big].\big(3\hat{\text{i}}+\hat{\text{j}}\big)=0$
$\Rightarrow(2-\lambda)3+(2+2\lambda)1+(3+\lambda)0=0$
$\Rightarrow6-3\lambda+2+2\lambda=0$
$\Rightarrow-\lambda+8=0$
$\Rightarrow\lambda=8$
Hence, the required value of $\lambda$ is 8.
View full question & answer→Question 253 Marks
Show that the four points having position vectors $6\hat{\text{i}}-7\hat{\text{j}},\ 16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},\ 3\hat{\text{j}}-6\hat{\text{k}},\ 2\hat{\text{i}}-5\hat{\text{j}}+10\hat{\text{k}}$ are coplanar.
AnswerLet the given four points be P, Q, R and S respectively. Three points are coplanar if the vectors $\overrightarrow{\text{PQ}},\ \overrightarrow{\text{PR}}\text{ and }\overrightarrow{\text{PS}}$ are coplanar. These vectors are coplanar if one of them can be expressed as a linear combination of the other two. So, let$\overrightarrow{\text{PQ}}=\text{x}\overrightarrow{\text{PR}}+\text{y}\overrightarrow{\text{PS}}$
$10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\text{x}\big(-6\hat{\text{i}}+10\hat{\text{j}}-6\hat{\text{k}}\big)+\text{y}\big(-4\hat{\text{i}}+2\hat{\text{j}}+10\hat{\text{k}}\big)$
$\Rightarrow10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\hat{\text{i}}\big(-6{\text{x}}-4\text{y}\big)+\hat{\text{j}}\big(10{\text{x}}+2{\text{y}}\big)+\hat{\text{k}}\big(-6\text{x}+10\text{y}\big)$
$\Rightarrow-6\text{x}-4\text{y}=10,\ 10\text{x}+2\text{y}=-12$ and $-6\text{x}+10\text{y}=-4$ [Equating coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ on both sides]
Solving the first of these three equations, we get x = -1 and y = -1. These values also satisfy the thied equation. Hence, the given four points are coplanar.
View full question & answer→Question 263 Marks
Find $\lambda$ when the projection of $\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ on $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$ is 4 units.
AnswerWe know
$\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is $\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
Given that
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}=4$
$\Rightarrow\frac{\big(\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}\big).\big(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big)}{\big|2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big|}$
$\Rightarrow\frac{2\lambda+6+12}{\sqrt{4+36+9}}=4$
$\Rightarrow\frac{2\lambda+18}{7}=4$
$\Rightarrow2\lambda+18=28$
$\Rightarrow2\lambda=10$
$\therefore\lambda=5$
View full question & answer→Question 273 Marks
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors:
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
AnswerGiven,
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjacent edges are $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$ Is equal to $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]$ Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}2&-3&4\\1&2&-1\\3&-1&-2 \end{vmatrix}$
$=2(-4-1)+3(2+3)+4(-1-6)$
$=-35$
volume of the parallelopiped $=\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|=|-35|=35$ cubic units.
View full question & answer→Question 283 Marks
Using vectors show that the points A(-2, 3, 5), B(7, 0, -1), C(-3. -2, -5) and D(3, 4, 7) are such that AB and CD intersect at the point P(1, 2, 3).
AnswerWe have,
$\overrightarrow{\text{AP}}=$ Position vector of P - Position vector of A
$\Rightarrow\overrightarrow{\text{AP}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)-\big(-2\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}\big)$
$=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
$\overrightarrow{\text{PB}}=$ Position vector of B - Position vector of P
$\Rightarrow\overrightarrow{\text{PB}}=\big(7\hat{\text{i}}-0\hat{\text{j}}-\hat{\text{k}}\big)-\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=6\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}$
Since $\overrightarrow{\text{PB}}=2\overrightarrow{\text{AP}}$. So, vectors $\overrightarrow{\text{PB}}$ and $\overrightarrow{\text{AP}}$ are collinear. But P is a point common to $\overrightarrow{\text{PB}}$ and $\overrightarrow{\text{AP}}$.
Hence P, A, B are collinear points.
Now, $\overrightarrow{\text{CP}}=\big(-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}\big)-\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\big(-4\hat{\text{i}}-4\hat{\text{j}}-8\hat{\text{k}}\big)$
$\overrightarrow{\text{PD}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)-\big(3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}\big)$
$=\big(-2\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}\big)$
Thus, $\overrightarrow{\text{CP}}=2\overrightarrow{\text{PD}}$
So the vectors $\overrightarrow{\text{CP}}$ and $\overrightarrow{\text{PD}}$ are collinear. But P is a common point to $\overrightarrow{\text{CP}}$ and $\overrightarrow{\text{PD}}$
Hence, C, P, D are collinear points.
Thus A, B, C, D and P are points such that A, P, B and C, P, D are two sets of collinear points.
Hence, AB and CD intersect at point P.
View full question & answer→Question 293 Marks
Write the position vector of the point which divdes the join of the points with position vectors $3\vec{\text{a}} - 2\vec{\text{b}} $ and $2\vec{\text{a}} + 3\vec{\text{b}} $ in the ratio 2 : 1.
AnswerSuppose R be the point which divdes the line joining the points with position vectors $3\vec{\text{a}} - 2\vec{\text{b}} $ and $2\vec{\text{a}} + 3\vec{\text{b}} $ in the ratio 2 : 1
And $\overrightarrow{\text{OA}} = 3\vec{\text{a}} - 2\vec{\text{b}} $ and $\overrightarrow{\text{OB}} = 2\vec{\text{a}} + 3\vec{\text{b}} $
Here, m : n = 2 : 1
Therefore, position vector $\overrightarrow{\text{OR}}$ is as follows:
$\overrightarrow{\text{OR}} = \frac{\text{m} \overrightarrow{\text{OB}} + \text{n} \overrightarrow{\text{OA}}}{\text{m + n}}$
$ = \frac{2\big(2\vec{\text{a}} + 3 \vec{\text{b}}\big) + 1 \big(3\vec{\text{a}} + 2 \vec{\text{b}}\big)}{2 + 1}$
$= \frac{7\vec{\text{a}} + 4\vec{\text{b}}}{3}$
$= \frac{7}{3} \vec{\text{a}} + \frac{4}{3} \vec{\text{b}}$
View full question & answer→Question 303 Marks
Let $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}.$ Then,
If $c_1= 1$ and $c_2= 2$, find $c_3$ which makes $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ coplanar.
AnswerIf $c_1 = 1$ and $c_2= 2$, then $\vec{\text{a}}=\hat{\text{i}}+\vec{\text{j}}+\vec{\text{k}},\vec{\text{b}}=\hat{\text{i}}$ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+\text{c}_3\hat{\text{k}}.$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar if $\big[\vec{\text{a}}\vec{\text{b}}{\text{c}}\big]=0.$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
$\Rightarrow\begin{vmatrix}1&1&1\\1&0&0\\1&2&\text{c}_3 \end{vmatrix}=0$
$\Rightarrow1(0-0)-1(\text{c}_3-0)+1(2-0)=0$
$\Rightarrow-{\text{C}}_3+2=0$
$\Rightarrow\text{C}_3=2$
View full question & answer→Question 313 Marks
$\text{Find}\ |\vec{a}|\ \text{and}\ \big|\vec{b}\big|,\text{if}\ (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=8\ \text{and}\ |\vec{a}|=8|\vec{b}|.$
Answer$\text{Given:}\ (\vec{a}+\vec{b}).(\vec{a}-\vec{b})=8\ \text{and}\ |\vec{a}|=8|\vec{b}|\ \ \ \ \ ....(\text{i})$
$\Rightarrow\ \ {\vec{a}.\vec{a}-\vec{a}.\vec{b}+\vec{b}.\vec{a}-\vec{b}.\vec{b}=8}{}$ $ \Rightarrow\ \ \big|\vec{a}\big|^2-\vec{a}.\vec{b}+\vec{a}.\vec{b}-\big|\vec{b}\big|^2=8$
$\Rightarrow\ \ \big|\vec{a}\big|^2-\big|\vec{b}\big|^2=8\ \ \ .....\text{(ii)}$
$\text{Putting}\ \big|\vec{a}\big|=8\big|\vec{b}\big|\ \text{in eq. (ii),}$
$ 64\bigg|\vec{b}\bigg|^2-\bigg|\vec{b}\bigg|^2=8 \ \ \Rightarrow\ \ \ (64-1)\bigg|\vec{b}\bigg|^2=8$
$\Rightarrow\ \ 63\bigg|\vec{b}\bigg|^2=8\ \ \Rightarrow\ \ \bigg|\vec{b}\bigg|^2=\frac{8}{63}$
$\Rightarrow\ \ \Big|\vec{b}\Big|=\sqrt{\frac{8}{63}}=\frac{2\sqrt{2}}{3\sqrt{7}}$
$\text{Putting}\ \Big|\vec{b}\Big|=\frac{2\sqrt{2}}{3\sqrt{7}}\ \text{in eq (i),}$
$\big|\vec{a}\big|=8\Big(\frac{2\sqrt{2}}{3\sqrt{7}}\Big)=\frac{16}{3}\sqrt{\frac{2}{7}}$
View full question & answer→Question 323 Marks
If $\overrightarrow{\text{PQ}}=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ and the coordinates of P are (1, -1, 2), find the coordinates of Q.
AnswerHere, $\overrightarrow{\text{PQ}}=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Position vector of $\text{P}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{PQ}}=$ Position vector of Q - Position vector of P
$3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}=$ Position vector of Q $-\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
Position vector of $\text{Q}=\big(\hat{3\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)+\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
$=4\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Coordinates of Q = (4, 1, 1)
View full question & answer→Question 333 Marks
Find the volume of the parallelopiped whose coterminous edges are represented by the vectore:
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Answer$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjacent edges
are $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ is equal to $\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|$ Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}1&1&1\\1&-1&1\\1&2&-1 \end{vmatrix}\\=1(1-2)-1(-1-1)+1(2+1)=4$
Volume of the parallelopiped $=\Big[\Big|\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|=|4|=4$ cubic units.
View full question & answer→Question 343 Marks
Dot product of a vector with vectore $\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ are respectively 4, 0 and 2. Find the vector.
AnswerLet $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ be the required vector.
Given that
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)=4$
$\Rightarrow\text{a}-\text{b+c}=4\dots(1)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)=0$
$\Rightarrow2\text{a}+\text{b}-3\text{c}=0\dots(2)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=2$
$\Rightarrow\text{a+b}+\text{c}=2\dots(3)$
Solving (1), (2) and (3), we get
$\text{a}=2,\text{b}=-1,\text{c}=1$
So, $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 353 Marks
A vector $\vec{\text{r}}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{\text{r}}$ is $2\sqrt3$, find $\vec{\text{r}}$.
AnswerLet l, m, n be the direction cosines of $\vec{\text{r}}$.
Now, $\vec{\text{r}}$ is inclined at equal angles to the three axes.
$\therefore\ \text{l}=\text{m}=\text{n}$ $[\alpha=\beta=\gamma\Rightarrow\cos\alpha=\cos\beta=\cos\gamma]$
So, $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ 3\text{l}^2=1$
$\Rightarrow\ \text{l}=\pm\frac{1}{\sqrt3}$
$\Rightarrow\ \text{l = m = n}=\pm\frac{1}{\sqrt3}$
We know that,
$\vec{\text{r}}=|\vec{\text{r}}|(\text{l}\hat{\text{i}}+\text{m}\hat{j}+\text{n}\hat{\text{k}})$
$\Rightarrow\ \vec{\text{r}}=2\sqrt3\Big(\pm\frac{1}{\sqrt3}\hat{\text{i}}\pm\frac{1}{\sqrt3}\hat{\text{j}}\pm\frac{1}{\sqrt3}\hat{\text{k}}\Big)$
$\Rightarrow\ \vec{\text{r}}=2\big(\pm\hat{\text{i}}\pm\hat{\text{j}}\pm\hat{\text{k}}\big)$
View full question & answer→Question 363 Marks
Write the number of vectors of unit length perpendicular to both the vectors $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$and $\vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}}.$
AnswerUnit vectors perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ are $\pm\Bigg(\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}\Bigg).$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&2\\0&1&1\end{vmatrix}=-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$\therefore$ Unit vectors perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ are $\pm\frac{-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{(-1)^2+(-2)^2+(2)^2}}=\pm\Big(-\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}+\frac{2}{3}\hat{\text{k}}\Big)$
Thus, there are two unit vectors perpendicular to the given vectors.
View full question & answer→Question 373 Marks
ABCD is a parallelogram and P is the point of intersection of its diagonals. If O is the origin of reference, show that $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=4\ \overrightarrow{\text{OP}}$.
Answer
Given a parallelogram ABCD and P is the point of intersection of its diagonals. We know the diagonals of a parallelogram, bisect eacg other. Therefore,
$\frac{\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}}2=\overrightarrow{\text{OP}}$
$\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}=2\ \overrightarrow{\text{OP}}\ \dots(1)$
and $\frac{\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}}2=\overrightarrow{\text{OP}}$
$\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}=2\ \overrightarrow{\text{OP}}\ \dots(2)$
Adding (1) and (2), We get,
$\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=4\ \overrightarrow{\text{OP}}$ View full question & answer→Question 383 Marks
Let $\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}.$ Find $\lambda$ such that $\vec{\text{a}}+\vec{\text{b}}$ is orthonal to $\vec{\text{a}}-\vec{\text{b}}.$
AnswerGive that
$\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}};\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$
$\therefore\vec{\text{a}}+\vec{\text{b}}=5\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}+\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}=6\hat{\text{i}}-2\hat{\text{j}}+(7+\lambda)\hat{\text{k}}$
and $\vec{\text{a}}-\vec{\text{b}}=5\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}-\big(\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}\big)=4\hat{\text{i}}+0\hat{\text{j}}+(7-\lambda)\hat{\text{k}}$
Given that $\vec{\text{a}}+\vec{\text{b}}$ is orthogonal to $\vec{\text{a}}-\vec{\text{b}}.$
$\Rightarrow\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow\big[6\hat{\text{i}}-2\hat{\text{j}}+(7+\lambda)\hat{\text{k}}\big].\big[4\hat{\text{i}}+0\hat{\text{j}}+(7-\lambda)\hat{\text{k}}\big]=0$
$\Rightarrow24+0+49-\lambda^2=0$
$\Rightarrow\lambda^2=73$
$\Rightarrow\lambda=\sqrt{73}$
View full question & answer→Question 393 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA.
AnswerSince, $\overrightarrow{\text{OA}}=\vec{\text{a}}$ and $\overrightarrow{\text{OB}}=\vec{\text{b}}$$\therefore\overrightarrow{\text{BA}}=\overrightarrow{\text{OA}}-\overrightarrow{\text{OB}}$
$=\vec{\text{a}}-\vec{\text{b}}$
Also, given that $\overrightarrow{\text{BC}}=1.5\ \overrightarrow{\text{BA}}$
$\Rightarrow\overrightarrow{\text{BC}}=1.5(\vec{\text{a}}-\vec{\text{b}})$
Or $\overrightarrow{\text{OC}}-\overrightarrow{\text{OB}}=1.5\vec{\text{a}}-1.5\vec{\text{b}}$
$\Rightarrow\overrightarrow{\text{OC}}=1.5\vec{\text{a}}-1.5\vec{\text{b}}+\vec{\text{b}}$ $[\because\overrightarrow{\text{OB}}=\vec{\text{b}}]$
$=1.5\vec{\text{a}}-0.5\vec{\text{b}}$
$=\frac{3\vec{\text{a}}-\vec{\text{b}}}{2}$
View full question & answer→Question 403 Marks
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors:
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjacent edges are $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$ Is equal to $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]$ Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}2&3&4\\1&2&-1\\3&-1&2 \end{vmatrix}$
$=2(4-1)-3(2+3)+4(-1-6)=-37$
volume of the parallelopiped $=\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|=|-37|=37$ cubic units.
View full question & answer→Question 413 Marks
Using vectors, find the value of k such that the points A(k, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.
AnswerLet the points are A(k, -10, 3), B(1, -1, 3) and C(3, 5, 4)
$\therefore\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})-(\text{k}\hat{\text{i}}-10\hat{\text{j}}+3\hat{\text{k}})$
$=(1+\text{k})\hat{\text{i}}+9\hat{\text{j}}$
and $\overrightarrow{\text{BC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OB}}$
$=(3\hat{\text{i}}-5\hat{\text{j}}+3\hat{\text{k}})-(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$
$=2\hat{\text{i}}+6\hat{\text{j}}$
If point A, B, C are collinear then there exists some scalar $'\lambda'$ such that
$\Rightarrow\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
$(1-\text{k})\hat{\text{i}}+9\hat{\text{j}}=\lambda(2\hat{\text{i}}+6\hat{\text{j}})$
Comparing coefficients we get $1-\lambda=2\lambda$ and $9=6\lambda$
$3-3\text{k}=9$
$\text{k}=-2$
View full question & answer→Question 423 Marks
In a triangle OAC, if B is the mid-point of side AC and $\overrightarrow{\text{OA}} = \vec{\text{a}}, \overrightarrow{\text{OB}} = \vec{\text{b}},$ then was is $\overrightarrow{\text{OC}}$?
AnswerIn $\triangle\text{OAC}, \overrightarrow{\text{OA}} = \vec{\text{a}} $ and $ \overrightarrow{\text{OB}} = \vec{\text{b}}.$
It is given that B is the mid-point of AC.
$\therefore$ Position vector of B $= \frac{\text{Positionn vector of A + Positionn vector of C}}{2}$
$\Rightarrow \overrightarrow{\text{OB}} = \frac{\overrightarrow{\text{OA}} + \overrightarrow{\text{OC}}}{2}$
$\Rightarrow \vec{\text{b}} = \frac{\vec{\text{a}}+\overrightarrow{\text{OC}}}{2}$
$\Rightarrow \vec{\text{a}} + \overrightarrow{\text{OC}} = 2\vec{\text{b}}$
$\Rightarrow \overrightarrow{\text{OC}} = 2\vec{\text{b}} - \vec{\text{a}}$ View full question & answer→Question 433 Marks
If the position vector $\vec{\text{a}}$ of a point (12, n) is such that $\big|\vec{\text{a}}\big|=13$, find the value (s) of n.
AnswerGiven a position vector $\vec{\text{a}}$ of a point (12, n) such that,
$\vec{\text{a}}=12\hat{\text{i}}+\text{n}\hat{\text{j}}$
Then,
$\big|\vec{\text{a}}\big|=\sqrt{12^2+\text{n}^2}$
Also, $\big|\vec{\text{a}}\big|=13$ (given)
Thus, we get,
$\sqrt{12^2+\text{n}^2}=13$
$\Rightarrow12^2+\text{n}^2=169$
$\Rightarrow\text{n}^2=169-144$
$\Rightarrow\text{n}^2=25$
$\Rightarrow\text{n}=\pm5$
View full question & answer→Question 443 Marks
If $|\vec{\text{a}}|=\sqrt{26,}\big|\vec{\text{b}}\big|=7$ and $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=35,$ find $\vec{\text{a}}.\vec{\text{b}}.$
Answer$\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta.\hat{\text{n}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\sin\theta||\hat{\text{n}}|$
$35=\sqrt{26.7}|\sin\theta|.1$
$\sin\theta=\frac{35}{\sqrt{26.5}}$
$\sin\theta=\frac{5}{\sqrt{26}}$
$\cos^2\theta=1-\sin^2\theta$
$=1-\Big(\frac{5}{\sqrt{26}}\Big)^2$
$=\frac{1}{1}-\frac{25}{26}$
$=\frac{26-25}{26}$
$=\frac{1}{26}$
$\cos\theta=\frac{1}{\sqrt{26}}$
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$=\sqrt{26}.7.\frac{1}{\sqrt{26}}$
$\vec{\text{a}}.\vec{\text{b}}=7$
View full question & answer→Question 453 Marks
Find the sine of the angle between the vectors $\vec{\text{a}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}.$
AnswerWe know that, angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ is given by
$\cos\theta=\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\vec{\text{a}}}||\vec{\text{b}}|}$
$=\frac{(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})\cdot(2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}})}{\sqrt{3^2+1^2+2^2}\sqrt{2^2+(-2)^2+4^2}}$
$\therefore\cos\theta=\frac{3}{\sqrt{21}}$
$\therefore\sin\theta=\sqrt{1-\cos^2\theta}$
$\sqrt{1-\frac{9}{21}}=\frac{2}{\sqrt{7}}$
View full question & answer→Question 463 Marks
Find the area of the parallelogram whose diagonals are:
$4\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$ and $-2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
AnswerArea of parallalogram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Here, $\vec{\text{d}}_1=4\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{d}}_2=-2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{d}_1}\times\vec{\text{d}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-1&-3\\-2&1&-2 \end{vmatrix}$
$=\hat{\text{i}}(2+3)-\hat{\text{j}}(-8-6)+\hat{\text{k}}(4-2)$
$=5\hat{\text{i}}+14\hat{\text{j}}+2\hat{\text{k}}$
$\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|=\sqrt{(5)^2+(14)^2+(2)^2}$
$=\sqrt{25+196+4}$
$=\sqrt{225}$
$=15$
Area of parallelgram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Area of parallelogram $=\frac{15}{2}\text{ sq. unit}$
View full question & answer→Question 473 Marks
If $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}\neq0,$ then show that $\vec{\text{a}}+\vec{\text{c}}=\text{m}\vec{\text{b}},$ where m is any scalar.
Answer$\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}$$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=-\vec{\text{c}}\times\vec{\text{b}}$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}=0$
$\Rightarrow\big(\vec{\text{a}}+\vec{\text{c}}\big)\times\vec{\text{b}}=0$ (using right distributive property)
Thus, $\vec{\text{a}}+\vec{\text{c}}$ is parallel to $\vec{\text{b}}.$
$\Leftrightarrow\vec{\text{a}}+\vec{\text{c}}=\text{m}\vec{\text{b}},$ for som scalar m.
View full question & answer→Question 483 Marks
If either vector $\vec{a}=\vec{0}\ \text{or}\ \vec{b}=\vec{0},\ \text{then}\ \vec{a}\cdot\vec{b}=0.$ But the converse need not be true. Justify your answer with an example.
AnswerCase I: $\ \text{Vector}\ \vec{a}=\vec{0}.$ Therefore by definition of zero vector, $\ \big|\vec{a}\big|=0\ \ ....\text{(i)}$$\therefore \ \vec{a}.\vec{b}=\big|\vec{a}\big|.\Big|\vec{b}\Big|\text{cos}\theta=0.\Big|\vec{b}\Big|\text{cos}\theta$ [Form eq. (i)]
$\Rightarrow\ \ \vec{a}.\vec{b}=0$
Case II: $\ \text{Vector}\ \vec{b}=\vec{0}.$ Therefore by definition of zero vector, $\ \big|\vec{b}\big|=0\ \ \ .....\text{(ii)}$
$\therefore\ \ \vec{a}.\vec{b}=\big|\vec{a}\big|.\Big|\vec{b}\Big|\text{cos}\theta=\vec{a}.0.\text{cos}\theta$ [Form eq. (ii)]
$\Rightarrow\ \ \vec{a}.\vec{b}=0$
But the converse is not true. Justification: $\\ \text{Let}\ \vec{a}=\hat{i}+\hat{j}+\hat{k}$$\text{Therefore},\ \big|\vec{a}\big|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}\neq0$
$\text{Therefore},\ \ \vec{a}\neq\vec{0}$
$\text{Again let}\ \ \ \vec{b}=\hat{i}+\hat{j}-2\hat{k}$
$\therefore\ \ \ \Big|\vec{b}\Big|=\sqrt{(1)^2+(1)^2+(-2)^2}=\sqrt{6}\neq0$
$\text{Therefore},\ \vec{b}\neq\vec{0}$
$\text{But}\ \ \ \vec{a}.\vec{b}=1(1)+1(1)+1(-2)=1+1+-2=0$
$\text{Hence, here}\ \vec{a}.\vec{b}=0,\ \text{but}\ \vec{a}\neq\vec{0}\ \text{and}\ \vec{b}\neq\vec{0}.$
View full question & answer→Question 493 Marks
Define $\vec{\text{a}}\times\vec{\text{b}}$ and prove that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta,$ where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerIf $\vec{\text{a}}$ and $\vec{\text{b}}$ are two non-parallel vectors, then the vectors product denoted by $\vec{\text{a}}\times\vec{\text{b}}$ is defined as $\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\hat{\text{ n}}.$
Here, $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ and $\hat{\text{n}}$ is the unit vector perpendicular to the plane of $\vec{\text{a}}$ and $\vec{\text{b}}$ such that $\vec{\text{a}},\vec{\text{b}}$ and $\hat{\text{n}}$ form a right
$\text{LHS}=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\times\frac{\cos\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\frac{\sin\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\tan\theta$
$=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta$
$=\text{RHS}$
Hence proved
View full question & answer→Question 503 Marks
If A, B, C, D are the points with position vectors $\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}},\ 2\hat{\text{i}}-3\hat{\text{k}},$ respectively, find the projection of $\overrightarrow{\text{AB}}$ along $\overrightarrow{\text{CD}}.$
AnswerWe have $\overrightarrow{\text{AB}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ \overrightarrow{\text{OB}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}},$
$\overrightarrow{\text{OC}}=\ 2\hat{\text{i}}-3\hat{\text{k}}$ and $\overrightarrow{\text{OD}}=3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\therefore\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$=\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
and $\overrightarrow{\text{CD}}=\overrightarrow{\text{OD}}-\overrightarrow{\text{OC}}$
$=(3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})-(2\hat{\text{i}}-3\hat{\text{k}})$
$=\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
Now the projection of $\overrightarrow{\text{AB}}$ along $\overrightarrow{\text{CD}}$
$=\overrightarrow{\text{AB}}.\frac{\overrightarrow{\text{CD}}}{|\overrightarrow{\text{CD}|}}$
$=\frac{(\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}})}{\sqrt{1^1+2^2+4^2}}$
$=\frac{1+4+16}{\sqrt{21}}=\frac{21}{\sqrt{21}}$
$=\sqrt{21}\text{ units}$
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