Question 12 Marks
If $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=3,$ find the projection of $\vec{\text{b}}$ on $\vec{\text{a}}.$
AnswerWe have
$|\vec{\text{a}}|=2$ and $\vec{\text{a}}.\vec{\text{b}}=3$
So, the projection of $\vec{\text{b}}$ on $\vec{\text{a}}$ is
$\Big(\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|}\Big)$
$=\frac{3}{2}$
View full question & answer→Question 22 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=4\hat{\text{i}}-4\hat{\text{j}}+7\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}$
$=(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-4\hat{\text{j}}+7\hat{\text{k}})$
$=(1)(4)+(-2).(-4)+(1)(7)$
$=4+8+7$
$=19$
$\vec{\text{a}}.\vec{\text{b}}=19$
View full question & answer→Question 32 Marks
Write the projection of the vector $7\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$ on the vector $2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}.$
AnswerLet $\vec{\text{a}}=7\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}};\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\Bigg(\frac{\vec{\text{a}}\vec{\text{b}}}{\big|\vec{\text{b}}\big|}\Bigg)$
$=\frac{\big(7\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}\big).\big(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big)}{\big|2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big|}$
$=\frac{14+6-12}{\sqrt{4+36+9}}$
$=\frac{8}{7}$
View full question & answer→Question 42 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=4$ and $\vec{\text{a}}.\vec{\text{b}}=1$
AnswerGiven that$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=4$ and $\vec{\text{a}}.\vec{\text{b}}=1\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=3^2+4^2-2(1)$ [using (1)]
$=9+16-2$
$=23$
$\therefore \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{23}$
View full question & answer→Question 52 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors of the same magnitude inclined at an angle of $60^\circ$ such that $\vec{\text{a}}.\vec{\text{b}}=8,$
write the value of their magnitude.
AnswerGiven that
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
and $\vec{\text{a}}$ and $\vec{\text{b}}$ are inclined at an angle of $60^\circ$
Also, given that
$\vec{\text{a}}.\vec{\text{b}}=8$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos60^\circ=8$
$\Rightarrow|\vec{\text{a}}||\vec{\text{a}}|\big(\frac{1}{2}\big)=8$
$\Rightarrow|\vec{\text{a}}|^2=16$
$\Rightarrow|\vec{\text{a}}|=4$
View full question & answer→Question 62 Marks
Find the value of $\theta\in(0,\frac{\pi}{2})$ for which vectors $\vec{\text{a}}=(\sin\theta)\hat{\text{i}}+(\cos\theta)\hat{\text{j}}$ and $\vec{\text{b}}=\hat{\text{i}}-\sqrt{3}\hat{\text{j}}+2\hat{\text{k}}$ are perpendicular.
AnswerWe have
$\vec{\text{a}}=(\sin\theta)\hat{\text{i}}+(\cos\theta)\hat{\text{j}}$
and
$\vec{\text{b}}=\hat{\text{i}}-\sqrt{3}\hat{\text{j}}+2\hat{\text{k}}$
It is given that the vectors are perpendicular.
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\sin\theta-\sqrt{3}\cos\theta=0$
$\Rightarrow\sin\theta=\sqrt{3}\cos\theta$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 72 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are vectors of equal magnitude, write the value of $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big).$
AnswerWE have
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|\dots(1)$
Now,
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)$
$=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$=|\vec{\text{a}}|^2-|\vec{\text{a}}|^2$ [using (1)]
$=0$
View full question & answer→Question 82 Marks
Find the value of $\lambda$ is the vectors $2\hat{\text{i}}+\lambda\hat{\text{j}}+3\hat{\text{k}}$ and $3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$ are perpendicular to each other.
AnswerGiven: $2\hat{\text{i}}+\lambda\hat{\text{j}}+3\hat{\text{k}}$ and $3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$ are perpendicular to each other.
So, their dot product is zero.
$\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+3\hat{\text{k}}\big).(3\text{i}+2\text{j}-4\text{k})$
$\Rightarrow6+2\lambda-12=0$
$\Rightarrow2\lambda-6=0$
$\Rightarrow\lambda=3$
View full question & answer→Question 92 Marks
If $\hat{\text{a}},\hat{\text{b}}$ are unit vectors such that $\hat{\text{a}}+\hat{\text{b}}$ is a unit vector, write the value of $\big|\hat{\text{a}}-\hat{\text{b}}\big|.$
AnswerGiven that $\hat{\text{a}}$ and $\hat{\text{b}}$ are unit vectors such that $\hat{\text{a}}+\hat{\text{b}}$ is a unit vector.
$\Rightarrow|\hat{\text{a}}|=\big|\hat{\text{b}}\big|=\big|\hat{\text{a}}+\hat{\text{b}}\big|=1\dots(1)$
Now,
$\big|\hat{\text{a}}+\vec{\text{b}}\big|=1$
Squaring both sides, we get
$|\vec{\text{a}}|^2+\big|\hat{\text{b}}\big|^2+2\hat{\text{a}}.\hat{\text{b}}=1$
$\Rightarrow1+1+2\hat{\text{a}}.\hat{\text{b}}=1$ [Form (1)]
$\Rightarrow\hat{\text{a}}.\hat{\text{b}}=\frac{-1}{2}\dots(2)$
Now,
$\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=|\text{a}|^2+\big|\hat{\text{b}}\big|^2-2\hat{\text{a}}.\hat{\text{b}}$
$=1+1-2\big(\frac{-1}{2}\big)=3$ [From (1) and (2)]
$\therefore\big|\hat{\text{a}}-\hat{\text{b}}\big|=\sqrt{3}$
View full question & answer→Question 102 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=2\hat {\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}} =4\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(2)^2+(-1)^2+(2)^{2}}=\sqrt{9}=3$
$\big|\vec{\text{b}}\big|=\sqrt{(4)^2+(4)^2+(-2)^{2}}=\sqrt{36}=6$
$\vec{\text{a}}.\vec{\text{b}}=8-4-4=0$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{0}{(3)(6)}=0$
$\Rightarrow\theta=\cos^{-1}(0)=\frac{\pi}{2}$
View full question & answer→Question 112 Marks
If $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+\hat{\text{k}},$ find the projection of $\vec{\text{a}}$ on $\vec{\text{b}}.$
AnswerWe have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
$=\frac{(\hat{\text{i}}-\hat{\text{j}}).(-\hat{\text{j}}+\hat{\text{k}})}{\big|-\hat{\text{j}}+\hat{\text{k}}\big|}$
$=\frac{0+1+0}{\sqrt{1+1}}$
$=\frac{1}{\sqrt{2}}$
View full question & answer→Question 122 Marks
Write the projection of $\vec{\text{r}}=3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}}$ on the coordinate axes.
AnswerWe have
$\vec{\text{r}}=3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}}$
Projection of $\vec{\text{r}}$ on x-axis $=\frac{\vec{\text{r}}.\hat{\text{i}}}{|\hat{\text{i}}|}=\frac{3}{1}=3$
Projection of $\vec{\text{r}}$ on y-axis $=\frac{\vec{\text{r}}.\hat{\text{j}}}{|\hat{\text{j}}|}=\frac{-4}{1}=-4$
Projection of $\vec{\text{r}}$ on z-axis $=\frac{\vec{\text{r}}.\hat{\text{k}}}{|\hat{\text{k}}|}=\frac{12}{1}=12$
View full question & answer→Question 132 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=4\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\Big).\Big(4\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}\Big)=0$
$\Rightarrow4\lambda-18+2=0$
$\Rightarrow4\lambda-16=0$
$\Rightarrow4\lambda=16$
$\Rightarrow\lambda=4$
View full question & answer→Question 142 Marks
Write the projection of $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ along the vector $\hat{\text{j}}.$
AnswerProjection of $\vec{\text{a}}$ on $\vec{\text{b}}=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
Projection of $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ along $\hat{\text{j}}$
$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big).\vec{\text{j}}}{|\vec{\text{j}}|}$
$=\frac{1}{1}$
$=1$
View full question & answer→Question 152 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}\big|=4,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=6,$ find the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.Given that
$\vec{\text{a}}.\vec{\text{b}}=6$
$\Rightarrow\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta=6$
$\Rightarrow(4)(3)\cos\theta=6$
$\Rightarrow12\cos\theta=6$
$\Rightarrow\cos\theta=\frac{6}{12}=\frac{1}{2}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{1}{2}\big)=\frac{\pi}{3}$
View full question & answer→Question 162 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three mutually perpendicular unit vectors, then prove that $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$
AnswerGiven that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.
So, $\big|\vec{\text{a}}\big|=1,\Big|\vec{\text{b}}\big|=1$ and $\big|\vec{\text{c}}\big|=1$
Since they are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|+2\vec{\text{a}}.\vec{\text{b}}+2\vec{{\text{b}}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}$
$=1+1+1+0+0+0$
$=3$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$
View full question & answer→Question 172 Marks
If $\vec{\text{a}}$ is a unit vector, then find $|\vec{\text{x}}|$ in each of the following.$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=12$
AnswerGiven that $\vec{\text{a}}$ is a unit vector.$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=12$
$\Rightarrow|\vec{\text{x}}|^2-|\vec{\text{a}}|^2=12$
$\Rightarrow|\vec{\text{x}}|^2-1^2=12$ [From (1)]
$\Rightarrow|\vec{\text{x}}|^2=13$
$\Rightarrow|\vec{\text{x}}|=\sqrt{13}$
View full question & answer→Question 182 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+\lambda\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(2\hat{\text{i}}+3 \hat{\text{j}}+4\hat{\text{k}}\Big).\Big(3\hat{\text{i}}+2\hat{\text{j}}-\lambda\hat{\text{k}}\Big)=0$
$\Rightarrow6+6-4\lambda=0$
$\Rightarrow12-4\lambda=0$
$\Rightarrow4\lambda=12$
$\Rightarrow\lambda=3$
View full question & answer→Question 192 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=\hat {\text{i}}+2\hat{\text{j}}-\hat{\text{k}},$ and $\vec{\text{b}} =\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(2)^2+(-1)^{2}}=\sqrt{6}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(-1)^2+(1)^{2}}=\sqrt{3}$
$\vec{\text{a}}.\vec{\text{b}}=1-2-1=-2$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-2}{\sqrt{6}\sqrt{3}}=\frac{-2}{\sqrt{18}}=\frac{-\sqrt{2}\times\sqrt{2}}{\sqrt{2}\times\sqrt{9}}=\frac{-\sqrt{2}}{3}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{-\sqrt{2}}{{3}}\Big)$
View full question & answer→Question 202 Marks
If $\vec{\text{a}}$ is a unit vector, then find $|\vec{\text{x}}|$ in each of the following.$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=8$
AnswerGiven that $\vec{\text{a}}$ is a unit vector.
$\Rightarrow|\vec{\text{a}}|=1\dots(1)$
$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=8$
$\Rightarrow|\vec{\text{x}}|^2-|\vec{\text{a}}|^2=8$
$\Rightarrow|\vec{\text{x}}|^2-1^2=8$ [From (1)]
$\Rightarrow|\vec{\text{x}}|^2=9$
$\Rightarrow|\vec{\text{x}}|=3$
View full question & answer→Question 212 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=\lambda\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(\lambda\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\Big).\Big(\hat{\text{i}}-1\hat{\text{j}}+3\hat{\text{k}}\Big)=0$
$\Rightarrow\lambda-3+6=0$
$\Rightarrow\lambda+3=0$
$\Rightarrow\lambda=-3$
View full question & answer→Question 222 Marks
Find the angle betwwen two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=1$
AnswerLet the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\theta,$ then
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}$
$=\frac{1}{3.3}$
$\cos\theta=\frac{1}{9}$
$\theta=\cos^{-1}\big(\frac{1}{9}\big)$
View full question & answer→Question 232 Marks
Find the angle between the vectora $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}.$
AnswerWe have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$|\vec{\text{a}}|=\sqrt{(1)^2+(-1)^2+(1)^2}=\sqrt{3}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2+(-1)^2}=\sqrt{3}$
and
$\vec{\text{a}}.\vec{\text{b}}=1-1-1=-1$
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{-1}{\sqrt{3}\sqrt{3}}=\frac{-1}{3}$
$\therefore\theta=\cos^-1\big(\frac{-1}{3}\big)$
View full question & answer→Question 242 Marks
If $\vec{\text{b}}$ is a unit vector such that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8,$ find $|\vec{\text{a}}|.$
AnswerGiven that $\vec{\text{b}}$ is a unit vector.
$\therefore\big|\vec{\text{b}}\big|=1$
And
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8$ (Given)
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow|\vec{\text{a}}|^2-1^2=8$
$\Rightarrow|\vec{\text{a}}|^2=9$
$\therefore|\vec{\text{a}}|=3$
View full question & answer→Question 252 Marks
If $\vec{\text{a}}.\vec{\text{a}}=0$ and $\vec{\text{a}}.\vec{\text{b}}=0,$ what can you conclude about the vector $\vec{\text{b}}$?
AnswerIt is given that $\vec{\text{a}}.\vec{\text{a}}=0$ and $\vec{\text{a}}.\vec{\text{b}}=0.$
Now,
$\vec{\text{a}}.\vec{\text{a}}=0\Rightarrow|\vec{\text{a}}|^2=0\Rightarrow|\vec{\text{a}}|=0$
$\therefore \vec{\text{a}}$ is a zero vector.
Hence, vector $\vec{\text{b}}$ satisfying $\vec{\text{a}}.\vec{\text{b}}=0.$ can be any vector
View full question & answer→Question 262 Marks
Write the value of p for which $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\text{p}\hat{\text{j}}+3\hat{\text{k}}$ are parallel vectors.
AnswerWe have$\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\text{p}\hat{\text{j}}+3\hat{\text{k}}$
Given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are parallel.
$\Rightarrow\vec{\text{a}}=\text{t}\vec{\text{b}}$ for some t.
$\Rightarrow3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}=\text{t}\big(\hat{\text{i}}+\text{p}\hat{\text{j}}+3\hat{\text{k}}\big)$
$\Rightarrow3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}=\text{t}\hat{\text{i}}+\text{pt}\hat{\text{j}}+3\text{t}\hat{\text{k}}$
Comparing both sides, we get
$3=\text{t,}2=\text{pt}$ and $9=3\text{t}$
$\Rightarrow\text{t}=3$ and $\text{pt}=2$
$\Rightarrow3\text{t}=2$
$\therefore\text{t}=\frac{2}{3}$
View full question & answer→Question 272 Marks
Find the cosine of the angle between the vectors $4\hat{\text{i}}-3\hat{\text{j}}+3\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}.$
AnswerLet, $\vec{\text{a}}=4\hat{\text{i}}-3\hat{\text{j}}+3\hat{\text{k}}$
and $\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$|\vec{\text{a}}|=\sqrt{(4)^2+(-3)^2+(3)^2}=\sqrt{34}$
$\big|\vec{\text{b}}\big|=\sqrt{(2)^2+(-1)^2+(-1)^2}=\sqrt{6}$
$\therefore\vec{\text{a}}.\vec{\text{b}}=8+3-3=8$
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{8}{\sqrt{34}\sqrt{6}}=\frac{8}{2\sqrt{51}}=\frac{4}{\sqrt{51}}$
View full question & answer→Question 282 Marks
If the vectors $3\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}$ and $18\hat{\text{i}}-12\hat{\text{j}}-\text{m}\hat{\text{k}}$ are parallel, find the value of m.
AnswerTHe given vectors are parallel.
$\therefore3\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}=\text{t}\big(18\hat{\text{i}}-12\hat{\text{j}}-\text{m}\hat{\text{k}}\big)$
$\Rightarrow3\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}=18\text{t}\hat{\text{i}}-12\text{t}\hat{\text{j}}-\text{t}\text{m}\hat{\text{k}}$
Comparing both sides, we get
$18\text{t}=3,-12\text{t}=-2,-4=\text{tm}$
$\Rightarrow\text{t}=\frac{1}{6}$
Substituting the value of m in -4 = -tm, we get
$-4=-\text{m}\big(\frac{1}{6}\big)$
$\therefore\text{m}=24$
View full question & answer→Question 292 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=2\hat {\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}} =\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.$\big|\vec{\text{a}}\big|=\sqrt{(2)^2+(-3)^2+(1)^{2}}=\sqrt{14}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2+(-2)^{2}}=\sqrt{6}$
$\vec{\text{a}}.\vec{\text{b}}=2-3-2=-3$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-3}{\sqrt{14}\sqrt{6}}=\frac{-3}{\sqrt{84}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{-3}{\sqrt{84}}\Big)$
View full question & answer→Question 302 Marks
Write the value of $\lambda$ so that vectora $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ are perpendicular to each other.
AnswerWe have
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
The given vectors are perpendicular. so, their dot product is zero.
$\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow5-2\lambda=0$
$\Rightarrow-2\lambda=-5$
$\Rightarrow\lambda=\frac{5}{2}$
View full question & answer→Question 312 Marks
Write the component of $\vec{\text{b}}$ along $\vec{\text{a}}.$
AnswerComponent of $\vec{\text{b}}$ on $\vec{\text{a}}$ is
$\Big\{\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|}\Big\}\hat{\text{a}}=\Bigg\{\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)}{|\vec{\text{a}}|^2}\Bigg\}\vec{\text{a}}=\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{a}}}{|\vec{\text{a}}|^2}$
View full question & answer→Question 322 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=8$
AnswerGiven that$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=8\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=2^2+5^2-2(8)$ [using (1)]
$=4+25-16$
$=13$
$\therefore \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{13}$
View full question & answer→Question 332 Marks
If the $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=\frac{2}{3}$ and $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector, then the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerLet the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ be $\theta.$
It is given that $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector.
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}\big|}=1$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta=1$
$\Rightarrow3\times\frac{2}{3}\times\sin\theta=1$
$\Rightarrow\sin\theta=\frac{1}{2}$
$\Rightarrow\theta=\frac{\pi}{6}$
Thus, the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6}.$
View full question & answer→Question 342 Marks
For any two non-zero vectors, write the value of $\frac{\big|\vec{\text{a}}+\vec{\text{b}}\big|^2+\big|\vec{\text{a}}-\vec{\text{b}}\big|^2}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}.$
AnswerWe have
$\frac{\big|\vec{\text{a}}+\vec{\text{b}}\big|^2+\big|\vec{\text{a}}-\vec{\text{b}}\big|^2}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=\frac{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}+|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=\frac{2\big(|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2\big)}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=2$
View full question & answer→Question 352 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=\hat {\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}} = \hat{\text{j}}+\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2}=\sqrt{2}$
$\vec{\text{a}}.\vec{\text{b}}=0-1+0=-1$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-1}{\sqrt{2}\sqrt{2}}=\frac{-1}{2}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{-1}{2}\big)=\frac{2\pi}{3}$
View full question & answer→Question 362 Marks
For what vaiue of $\lambda$ are the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ perpendicular to each other?
AnswerWe know
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Given, $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular. so, their dot product is zero.
$\Rightarrow\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow-2\lambda+5=0$
$\therefore\lambda=\frac{5}{2}$
View full question & answer→Question 372 Marks
If $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are mutually perpendicular unit vectors, write the value of $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|.$
AnswerGiven that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.
So, $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=1$ and $|\vec{\text{c}}|=1\dots(1)$
Since they are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0\dots(2)$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|+|\vec{\text{c}}|^2+2\vec{\text{a}}.\vec{\text{b}}+2\vec{\text{b}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}$
$=1+1+1+0+0+0$ [using (1) and (2)]
$=3$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$
View full question & answer→Question 382 Marks
If $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=2,$ find $\big|\hat{\text{a}}-\hat{\text{b}}\big|.$
View full question & answer→Question 392 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\vec{\text{a}}.\vec{\text{b}}=6,|\vec{\text{a}}|=3$ and $\big|\vec{\text{b}}\big|=4.$ write the projection of $\vec{\text{a}}$ on $\vec{\text{b}}.$
AnswerWe have
$\vec{\text{a}}.\vec{\text{b}}=6$ and $\big|\vec{\text{b}}\big|=4$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
$=\frac{6}{4}$
$=\frac{3}{2}$
View full question & answer→Question 402 Marks
Write the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes $\sqrt{3}$ and 2 respectively if $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}.$
AnswerLet $\theta$ be the angle between$\vec{\text{a}}$ and$\vec{\text{b}}.$
Given,
$|\vec{\text{a}}|=\sqrt{3};\big|\vec{\text{b}}\big|=2;\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{6}=(\sqrt{3})(2)\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{6}}{2\sqrt{3}}=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=\frac{\pi}{4}$
View full question & answer→Question 412 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=3\hat {\text{i}}-2\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}} =4\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(3)^2+(-2)^2+(-6)^{2}}=\sqrt{49}=7$
$\big|\vec{\text{b}}\big|=\sqrt{(4)^2+(1)^2+(8)^{2}}=\sqrt{81}=9$
$\vec{\text{a}}.\vec{\text{b}}=12+2-48=-34$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-34}{(7)(9)}=\frac{-34}{63}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{-34}{63}\big)$
View full question & answer→Question 422 Marks
Write the projection of the vector $\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}$ on the vector $2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.$
AnswerWe know that projection of $\vec{\text{a}}$ on $\vec{\text{b}}=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}.$
Let $\vec{\text{a}}=\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.$
$\therefore$ projection of $\vec{\text{a}}$ on $\vec{\text{b}}$
$=\frac{\big(\hat{\text{i}}+3\hat{\text{j}}+7\vec{\text{k}}\big).\big(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.\big)}{\big|2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.\big|}$
$=\frac{1\times2+3\times(-3)+7\times6}{\sqrt{2^2+(-3)^2+6^2}}$
$=\frac{2-9+42}{\sqrt{49}}$
$=\frac{35}{7}$
$=5$
View full question & answer→Question 432 Marks
If two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=1$ and $\vec{\text{a}}.\vec{\text{b}}=1,$ then find the value of $\big(3\vec{\text{a}}-5\vec{\text{b}}\big).\big(2\vec{\text{a}}+7\vec{\text{b}}\big).$
Answer$\big(3\vec{\text{a}}-5\vec{\text{b}}\big).\big(2\vec{\text{a}}+7\vec{\text{b}}\big)$
$=3\vec{\text{a}}.2\vec{\text{a}}+3\vec{\text{a}}.7\vec{\text{b}}-5\vec{\text{b}}.2\vec{\text{a}}-5\vec{\text{b}}.7\vec{\text{b}}$
$=6\vec{\text{a}}.\vec{\text{a}}+21\vec{\text{a}}.\vec{\text{b}}-10\vec{\text{a}}.\vec{\text{b}}-35\vec{\text{b}}.\vec{\text{b}}$
$=6|\vec{\text{a}}|^2+11\vec{\text{a}}.\vec{\text{b}}-35\big|\vec{\text{b}}\big|^2$
$=6\times2^2+11\times1-35\times1^2$
$=35-35$
$=0$
View full question & answer→Question 442 Marks
Find the manitude of two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ that are of the same magnitude, are inclined at 60° and whose scalar product is $\frac{1}{2}.$
AnswerGiven that the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $30^{\circ}.$
Also,
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|;\vec{\text{a}}.\vec{\text{b}}=\frac{1}{2}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\frac{1}2{}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos60$
$\Rightarrow\frac{1}{2}=|\vec{\text{a}}|^2\big(\frac{1}{2}\big)$
$\Rightarrow|\vec{\text{a}}|^2=1$
$\Rightarrow|\vec{\text{a}}|=1$
$\therefore|\vec{\text{a}}|=\big|\vec{\text{b}\big|}=1$
View full question & answer→Question 452 Marks
Write a vector satisfying $\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1.$
AnswerLet $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{a}}.\hat{\text{i}}=\text{a}_1$
$\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\text{a}_1+\text{a}_2$
$\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{a}_1+\text{a}_2+\text{a}_3$
Given that
$\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1$
$\Rightarrow\text{a}_1=\text{a}_1+\text{a}_2=\text{a}_1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_1+\text{a}_2=1;\text{a}_1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;1+\text{a}_2=1;1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_2=0;1+0+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_2=0;\text{a}_3=0$
So, $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\hat{\text{i}}$
View full question & answer→Question 462 Marks
If $\vec{\text{a}}.\vec{\text{a}}=0$ and $\vec{\text{a}}.\vec{\text{b}}=0,$ what can you conclude about the vector $\vec{\text{b}}?$
AnswerGiven that $\vec{\text{a}}.\vec{\text{a}}=0$
$\Rightarrow|\vec{\text{a}}|^2=0$
$\Rightarrow|\vec{\text{a}}|=0\dots(1)$
Also, given that
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=0$ (where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$)
$\Rightarrow0\big|\vec{\text{b}}\big|\cos\theta=0$ [From (1)]
$\Rightarrow0=0$
So, it means that for any vector $\vec{\text{b}},$ the given equation $\vec{\text{a}}.\vec{\text{b}}=0$ is satisfid.
View full question & answer→Question 472 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ write when $\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{a}}-\vec{\text{b}}\big|$ holds.
AnswerGiven that
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{a}}-\vec{\text{b}}\big|$
Squaring both sides, we get
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=\big|\vec{\text{a}}-\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$\Rightarrow4\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
⇒ $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendiculalr.
View full question & answer→Question 482 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, find the angle between $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}.$
AnswerWe have
$|\vec{\text{a}}|=1$ and $\big|\vec{\text{b}}\big|=1\dots(1)$
Now,
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$=1^2-1^2$ [using (1)]
$=0$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}$ are perpendicular.
$\therefore$ angle between $\big(\vec{\text{a}}+\vec{\text{b}}\big)$ and $\big(\vec{\text{a}}-\vec{\text{b}}\big)$ is 90°.
View full question & answer→Question 492 Marks
Write the value of $\big(\vec{\text{a}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{a}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{a}}.\hat{\text{k}}\big)\hat{\text{k}},$ where $\vec{\text{a}}$ is any vector.
AnswerLet $\vec{\text{a}}=\text{a}_1\vec{\text{i}}+\text{a}_2\vec{\text{j}}+\text{a}_3\vec{\text{k}}$
Now,
$\big(\vec{\text{a}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{a}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{a}}.\hat{\text{k}}\big)\hat{\text{k}}$
$=\text{a}_1\vec{\text{i}}+\text{a}_2\vec{\text{j}}+\text{a}_3\vec{\text{k}}$
$=\vec{\text{a}}$
View full question & answer→Question 502 Marks
Find the projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ if $\vec{\text{a}}.\vec{\text{b}}=8$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}.$
AnswerWe have
$\vec{\text{a}}.\vec{\text{b}}=8$ 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
$\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{b}}|}$
$=\frac{8}{\sqrt{4+36+9}}$
$=\frac{8}{7}$
View full question & answer→Question 512 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}=(\hat{\text{j}}+2\hat{\text{k}}).(2\hat{\text{i}}+\hat{\text{k}})$
$=(0\times\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})(2\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}})$
$=(0)(2)+(1)(0)+(2)(1)$
$=0+0+2$
$\vec{\text{a}}.\vec{\text{b}}=2$
View full question & answer→Question 522 Marks
For any two vectore $\vec{\text{a}}$ and $\vec{\text{b}}$, show that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0\Leftrightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|.$
AnswerWe have
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=0$
$\Rightarrow|\vec{\text{a}}|^2=\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
View full question & answer→Question 532 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=5\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\Big).\Big(5\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}\Big)=0$
$\Rightarrow5\lambda-18+2=0$
$\Rightarrow5\lambda-16=0$
$\Rightarrow5\lambda=16$
$\Rightarrow\lambda=\frac{16}{5}$
View full question & answer→Question 542 Marks
What is the angle between vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes 2 and $\sqrt{3}$ respectively? Given $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}.$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Given that
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=\sqrt{3}$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{3}=(2)(\sqrt{3})\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{3}}{2\sqrt{3}}$
$\Rightarrow\cos\theta=\frac{1}2{}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{1}2{}\big)=\frac{\pi}{3}$
View full question & answer→Question 552 Marks
If the vectors $3\hat{\text{i}}+\text{m}\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}-8\hat{\text{k}}$ are orthonal, find m.
AnswerIt is given that the vectors are othgonal. so, their dot product is zero.
$\big(3\hat{\text{i}}+\text{m}\hat{\text{j}}+\hat{\text{k}}\big).\big(2\hat{\text{i}}-\hat{\text{j}}-8\hat{\text{k}}\big)=0$
$\Rightarrow6-\text{m}-8=0$
$\Rightarrow-\text{m}-2=0$
$\Rightarrow\text{m}=-2$
View full question & answer→Question 562 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are mutually perpendicular unit vectors, write the value of $\big|\vec{\text{a}}+\vec{\text{b}}\big|.$
Answer$\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors and they are perpendicular.
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1;\vec{\text{a}}.\vec{\text{b}}=0\dots(1)$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}$
$=1+1+2(0)$ [using (1)]
$=2$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{20}$
View full question & answer→Question 572 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big).=0,$ find the relation betwen the magnitudes of $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerGiven that$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=0$
$\Rightarrow|\vec{\text{a}}|^2=\big|\vec{\text{b}}\big|^2$
$\therefore|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
View full question & answer→Question 582 Marks
For any two vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ write when $\big|\vec{\text{a}}+\vec{\text{b}}\big|=|\vec{\text{a}}|+\big|\vec{\text{b}}\big|$ holds.
AnswerGiven that $\big|\vec{\text{a}}+\vec{\text{b}}\big|=|\vec{\text{a}}|+\big|\vec{\text{b}}\big|$ Squaring both sides, we get $\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=\big(|\vec{\text{a}}|+\big|\vec{\text{b}}\big|\big)^2$ $\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ $\Rightarrow\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ $\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ (where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$) $\Rightarrow\cos\theta=1$ $\Rightarrow\theta=0^{\circ}$$\Rightarrow\vec{\text{a}}$ and $\vec{\text{b}}$ are parallel.
View full question & answer→Question 592 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}=(\hat{\text{j}}-\hat{\text{k}}).(2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}})$
$=(0\times\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})(2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}})$
$=(0)(2)+(1)(3)+(-1)(-2)$
$=0+3+2$
$\vec{\text{a}}.\vec{\text{b}}=5$
View full question & answer→Question 602 Marks
If $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+2\hat{\text{k}},$ find $\big(\vec{\text{a}}-2\vec{\text{b}}\big).\big(\vec{\text{a}}+\vec{\text{b}}\big).$
AnswerWe have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{a}}-2\vec{\text{b}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)-2\big(-\hat{\text{j}}+2\hat{\text{k}}\big)\\=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{j}}-4\hat{\text{k}}=\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{a}}+\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}-\hat{\text{j}}+2\hat{\text{k}}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$\big(\vec{\text{a}}-2\vec{\text{b}}\big).\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$=\big(\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$
$=1-2-8$
$=-9$
View full question & answer→Question 612 Marks
Find the angle betwwen two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if$|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=2$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
AnswerWe have,
$|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=2$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$. then
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{6}}{\sqrt{3}\times2}$
$=\frac{1}{\sqrt{2}}$
$\theta=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$
$=\frac{\pi}{4}$
View full question & answer→Question 622 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=4$
AnswerGiven that$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=4\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=2^2+3^2-2(4)$ [using (1)]
$=4+9-8$
$=5$
$\therefore \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{5}$
View full question & answer→Question 632 Marks
For what of $\lambda$ are the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ perpendicular to each other?
AnswerWe have
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular.
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow5-2\lambda=0$
$\therefore\lambda=\frac{5}{2}$
View full question & answer→