Question 512 Marks
Write the direction cosines of the vectors $-2\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}$.
AnswerGiven: $-2\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}$
Then, its direction cosines are:
$\frac{-2}{\sqrt{(-2)^2+1^2+(-5)^2}},\frac{1}{\sqrt{(-2)^2+1^2+(-5)^2}},\frac{-5}{\sqrt{(-2)^2+1^2+(-5)^2}}$ or, $\frac{-2}{\sqrt{30}},\frac{1}{\sqrt{30}},\frac{-5}{\sqrt{30}}$
View full question & answer→Question 522 Marks
Write the value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big).$
Answer$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big)$
$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}.\hat{\text{k}}$
$=|\hat{\text{i}}|^2+|\hat{\text{j}}|^2+|\hat{\text{k}}|^2$
$=1+1+1$
$=3$
View full question & answer→Question 532 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$, find a vecctor of magnitude 6 units which is parallel to the vector $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$.
AnswerWe have, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$Then,
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)\\-\big(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)+3\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$ $=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ $\therefore$ A unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$ is $\frac{2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}}{\big|2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\big|}=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{1^2+(-2)^2+2^2}}$ $=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt9}$ $=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}3$ Hence, Required vector $=\frac{6}3\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$ $=2\hat{\text{i}}-4\hat{\text{j}}+4\hat{\text{k}}$
View full question & answer→Question 542 Marks
Find the magnitude of the vector $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$$\therefore$ Magnitude of the vector $=\big|\vec{\text{a}}\big|=\sqrt{2^2+3^2+(-6)^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
View full question & answer→Question 552 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 562 Marks
vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=\frac{2}{3}$ and $\big(\vec{\text{a}}\times\vec{\text{b}}\big)$ is a unit vector. write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerGiven: $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector.
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1\dots(1)$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
From (1), we get
$1=(\sqrt{3})\Big(\frac{2}{3}\Big)\sin\theta$
$\Rightarrow\sin\theta=\frac{\sqrt{3}}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 572 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 582 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors such that $\vec{\text{a}}\times\vec{\text{b}}$ is also a unit vector, 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:
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1$
$|\vec{\text{a}}|=1$
$|\vec{\text{b}}|=1$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow1=(1)(1)\sin\theta$
$\Rightarrow\sin\theta=1$
$\Rightarrow\theta=\frac{\pi}{2}$
View full question & answer→Question 592 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors of magnitudes 3 and $\frac{\sqrt{2}}{3}$ repectively such that $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector. write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerWrite the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
It is given that $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector.
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow1=(3)\Big(\frac{\sqrt{2}}{3}\big)\sin\theta$
$\Rightarrow\sin\theta=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=45^\circ,135^\circ$
View full question & answer→Question 602 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 612 Marks
Find the volume of the parallelopiped with its edges represented by the vectors
$\hat{\text{i}}+\hat{\text{j}},\hat{\text{i}}+2\hat{\text{j}}$ and $\hat{\text{i}}+\hat{\text{j}}+\pi{\text{k}}.$
AnswerLet:
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}$
$\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\pi\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjecent edges are $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ is equal to $\big|\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]\big|.$
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\begin{vmatrix}1&1&0\\1&2&0\\1&1&\pi \end{vmatrix}$
$=1(2\pi-0)-1(\pi-0)+0(1-2)$
$=2\pi-\pi=\pi$
$\therefore$ volume $=\big|\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]\big|=|\pi|=\pi$ cubic units
View full question & answer→Question 622 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are position vectors of the vertices A, B and C respectively, of a triangle ABC, write the value of$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}$.
AnswerGiven: $\vec{\text{a}},\vec{\text{b}}\text{ and }\vec{\text{c}}$ are the position vectors of A, B and C respectively. Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\overrightarrow{\text{BC}}=\vec{\text{c}}-\vec{\text{b}}$
$\overrightarrow{\text{CA}}=\vec{\text{a}}-\vec{\text{c}}$
Consider,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}\\=\vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{b}}+\vec{\text{a}}-\vec{\text{c}}$
$=\vec0$
View full question & answer→Question 632 Marks
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ represent two adjacent sides of a parallelogram, then write vectors representing its diagonals.
AnswerLet $\vec{\text{a}}\text{ and }\vec{\text{b}}$ represents two adjacent sides of a parallelogram ABCD.
$\therefore$ AB = DC and AD = BC
$\Rightarrow\ \overrightarrow{\text{DC}}=\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\Rightarrow\ \overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}=\vec{\text{b}}$
In $\triangle\text{ABC}$
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
$\Rightarrow\ \vec{\text{a}}+\vec{\text{b}}=\overrightarrow{\text{AC}}$
In $\triangle\text{ABD}$
$\ \overrightarrow{\text{AD}}+\overrightarrow{\text{DB}}=\overrightarrow{\text{AB}}$
$\Rightarrow\ \vec{\text{b}}+\overrightarrow{\text{DB}}=\vec{\text{a}}$
$\Rightarrow\ \overrightarrow{\text{DB}}=\vec{\text{a}}-\vec{\text{b}}$
View full question & answer→Question 642 Marks
Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.
AnswerSuppose the vector makes equal angle with the coordinate axis.
Then, its direction cosines are . Therefore,
$\text{l}=\text{m}=\text{n}.$
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow3\text{l}^2=1$
$\Rightarrow\text{l}^2=\frac{1}{3}$
$\Rightarrow\text{l}=\frac{1}{\sqrt{3}}$
Hence, direction cosines are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
View full question & answer→Question 652 Marks
Find the value of 'p' for which the vectors $3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}$ and $\hat{\text{i}}-2\text{p}\hat{\text{j}}+3\hat{\text{k}}$ are parallel.
AnswerLet $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\text{p}\hat{\text{j}}+3\hat{\text{k}}$ be the two given vectors.
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are parallel, then
$\vec{\text{b}}=\lambda\vec{\text{a}}$ for some scalar $\lambda$
$\therefore\ \hat{\text{i}}-2\text{p}\hat{\text{j}}+3\hat{\text{k}}=\lambda\big(3\hat{\text{i}}-2\hat{\text{j}}+9\hat{\text{k}}\big)$
$\Rightarrow\ \hat{\text{i}}-2\text{p}\hat{\text{j}}+3\hat{\text{k}}=3\lambda\hat{\text{i}}+2\lambda\hat{\text{j}}+9\lambda\hat{\text{k}}$
$\Rightarrow\ 1=\lambda3$ and $-2\text{p}=2\lambda$ $\big($ Equating coefficients of $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}\big)$
$\Rightarrow\ \text{p}=-\lambda=-\frac{1}3$
Thus, the value of p is $-\frac{1}3$.
View full question & answer→Question 662 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 672 Marks
If $\vec{\text{a}}$ is a vector and m is a scalar such that m $\vec{\text{a}}=\vec0$, then what are the alternatives for m and $\vec{\text{a}}$?
AnswerGiven: $\vec{\text{a}}$ is a vector and m is a scalar such that, $\text{m}\vec{\text{ a}}=\vec0$
Then either $\text{m}=0\text{ or, } \vec{\text{a}}=\vec0$
View full question & answer→Question 682 Marks
Find a vector in the direction of $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$, which has magnitude of 6 units.
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{2^2+(-1)^2+2^2}$
$=\sqrt{4+1+4}$
$=\sqrt{9}$
$=3$
$\therefore$ Required Vector $=6\times\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=6\times\frac{\big(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)}3$
$=4\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
View full question & answer→Question 692 Marks
If $\vec{\text{a}}=\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+3\hat{\text{k},}$ find $\big|\vec{\text{a}}\times\vec{\text{b}}\big|.$
AnswerIf $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{b}_1\hat{\text{j}}+\text{c}_1\hat{\text{k}}$ and
$\vec{\text{b}}=\text{a}_2\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{c}_2\hat{\text{k}},$ then
$\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}_2\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&3&-2\\-1&0&3 \end{vmatrix}$
$=\hat{\text{i}}(9-0)-\hat{\text{j}}(3-2)+\hat{\text{k}}(0+3)$
$\vec{\text{a}}\times\vec{\text{b}}=9\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{(9)^2+(-1)^2+(3)^2}$
$=\sqrt{81+1+9}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{91}$
View full question & answer→Question 702 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 712 Marks
Write the value of $\big[2\hat{\text{i }} 3\hat{\text{j }}4\hat{\text{k}}\big].$
AnswerWe have
$\big[2\hat{\text{i }}3\hat{\text{j }}4\hat{\text{k}}\big]$
$=\big(2\hat{\text{i}}\times3\hat{\text{j}}\big).4\hat{\text{k}}$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$=6\hat{\text{k}}.4\hat{\text{k}}$
$=24$
View full question & answer→Question 722 Marks
Write the expression for the area of the parallelogram having $\vec{\text{a}}$ and $\vec{\text{b}}$ as its diagonals.
AnswerGiven: $\vec{\text{a}}$ and $\vec{\text{b}}$ are diagonals of a parallelogram.
Area of the parallelogram $=\frac{1}{2}\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
View full question & answer→Question 732 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 742 Marks
Find the unit vector parallel to the vector $\hat{\text{i}}+\sqrt3\hat{\text{j}}$.
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\sqrt3\hat{\text{j}}$
Then, $\big|\vec{\text{a}}\big|=\sqrt{1^2+\big(\sqrt3\big)^2}$
$=\sqrt{1+3}$
$=\sqrt4$
$=2$
Unit vector parallel to $\vec{\text{a}}=\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}{2}\big(\hat{\text{i}}+\sqrt3\hat{\text{j}}\big)=\frac{1}2\hat{\text{i}}+\frac{\sqrt3}2\hat{\text{j}}$
View full question & answer→Question 752 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 762 Marks
Find the value of x for which $\text{x}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ is a unit vector.
AnswerWe have, $\text{x}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ is a unit vector.
$\therefore\ \sqrt{\text{x}^2+\text{x}^2+\text{x}^2}=1$
$\Rightarrow\sqrt3|\text{x}|=1$
$\Rightarrow|\text{x}|=\frac{1}{\sqrt3}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt3}$
View full question & answer→Question 772 Marks
For any three vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ write the value of $\vec{\text{a}}\times\big(\vec{\text{b}}+\vec{\text{c}}\big)+\vec{\text{b}}\times\big(\vec{\text{c}}+\vec{\text{a}}\big)+\vec{\text{c}}\times\big(\vec{\text{a}}+\vec{\text{b}}\big).$
Answer$\vec{\text{a}}\times\big(\vec{\text{b}}+\vec{\text{c}}\big)+\vec{\text{b}}\times\big(\vec{\text{c}}+\vec{\text{a}}\big)+\vec{\text{c}}\times\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\big(\vec{\text{c}}\times\vec{\text{a}}\big)+\big(\vec{\text{c}}\times\vec{\text{b}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\big(\vec{\text{a}}\times\vec{\text{c}}\big)-\big(\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\vec{0}$
View full question & answer→Question 782 Marks
Write the value of $\big[\hat{\text{i}}-\hat{\text{j }}\hat{\text{j}}-\hat{\text{k }}\hat{\text{k}}-\hat{\text{i}}\big].$
AnswerWe have
$\big[\hat{\text{i}}-\hat{\text{j }}\hat{\text{j}}-\hat{\text{k }}\hat{\text{k}}-\hat{\text{i}}\big]=\big[\big(\hat{\text{i}}-\hat{\text{j}}\big)\times\big(\hat{\text{j}}-\hat{\text{k}}\big)\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$\big[\big(\hat{\text{i}}\times\hat{\text{j}}\big)-\big(\hat{\text{i}}\times\hat{\text{k}}\big)-\big(\hat{\text{j}}\times{\hat{\text{j}}}\big)+\big(\hat{\text{j}}\times\hat{\text{k}}\big)\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$=\big[\hat{\text{k}}+\hat{\text{j}}+\hat{\text{i}}\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$=\big[\big(\hat{\text{k}}.\hat{\text{k}}\big)-\big(\hat{\text{k}}.\hat{\text{i}}\big)+\big(\hat{\text{j}}.\hat{\text{k}}\big)-\big(\hat{\text{j}}.\hat{\text{i}}\big)+\big(\hat{\text{i}}.\hat{\text{k}}\big)-\big(\hat{\text{i}}.\hat{\text{i}}\big)\big]$
$=1-0+0-0+0-1=0$
View full question & answer→Question 792 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3}$ and $\vec{\text{a}}.\vec{\text{b}}=1,$ find the angle between.
Answer$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3}$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta=\sqrt{3}\dots(1)$
$\vec{\text{a}}.\vec{\text{b}}=1$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=1\dots(2)$
Dividing (1) by (2), we get
$\frac{|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta}=\sqrt{3}$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\theta=60^\circ$
View full question & answer→Question 802 Marks
Find the components along the coordinate axis of the position vector of the following point:R(-11, -9)
AnswerHere, R = (-11, -9)
Position vector of $\text{R}=-11\hat{\text{i}}-9\hat{\text{j}}$
Component of R along x-axis $=-11\hat{\text{i}}$
Component of R along x-axis $=-9\hat{\text{j}}$
View full question & answer→Question 812 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 822 Marks
Write two different vectors having same magnitude.
AnswerConsider $\vec{a}=(\hat{i}-2\hat{j}+3\hat{k})\ \text{and}\ \vec{b}=(2\hat{i}+\hat{j}-3\hat{k}).$ It can be observed that $\Big|\vec{a}\Big|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+ 4+9}=\sqrt{14}\ \text{and}$$\Big|\vec{b}\Big|=\sqrt{2^2+1^2+(-3)^2}=\sqrt{4+ 1+9}=\sqrt{14}.$
Hence, $\vec{a}\ \text {and}\ \vec{b}$ are two different vectors having the same magnitude. The vectors are different because they have different directions.
View full question & answer→Question 832 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$, write unit vectors parallel to $\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$Now, $\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{j}}+\hat{\text{k}}-2\hat{\text{k}}-2\hat{\text{i}}$
$=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ Unit vector parallel to $\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}=\frac{-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}{\sqrt{(-1)^2+2^2+(-1)^2}}$ $$$=\frac{-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}{\sqrt6}$
View full question & answer→Question 842 Marks
$\text{Find}\ \lambda\ \text{and}\ \mu\ \text{if}\ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=\vec{0}.$
Answer$\text{Given:}\ \ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=0$
$\Rightarrow\ \ \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&6&27\\1&\lambda&\mu\end{vmatrix}=\vec{0}$
Expanding along first row,
$\hat{i}(6\mu-27\lambda)-\hat{j}(2\mu-27)+\hat{k}(2\lambda-6)$ $=\vec{0}=0\hat{i}+0\hat{j}+0\hat{k}$
Comparing the coefficients of $\hat{i}, \hat{j},\hat{k}$ on both sides, we have
$6\mu-27\lambda=0\ \ \ \ \ .....\text{(i)}$
$2\mu-27=0\ \ \ \ \ \ .....\text{(ii)}$
$\text{And}\ \ 2\lambda-6=0\ \ \ \ \ \ .....\text{(iii)}$
$\text{From eq. (ii),}\ \ 2\mu-27=0 \ \Rightarrow\ \ \mu=\frac{27}{2}$
$\text{From eq. (iii),}\ \ 2\lambda-6=0\ \Rightarrow\ \ \lambda=\frac{6}{2}=3$
Putting the values of $\mu\ \text{and}\ \lambda$ in eq. (i),
$6\bigg(\frac{27}{2}\bigg)-27(3)=0\ $ $ \Rightarrow\ 3(27)-27(3)=0\ \Rightarrow\ \ 0=0$
$\text{Therefore,}\ \ \mu=\frac{27}{2}\ \text{and}\ \lambda=\frac{6}{2}=3$
View full question & answer→Question 852 Marks
If $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+2\hat{\text{k}}$, write a unit vector along the vector $3\vec{\text{a}}-2\vec{\text{b}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+2\hat{\text{k}}$
Therefore,
$3\vec{\text{a}}-2\vec{\text{b}}=3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{j}}-4\hat{\text{k}}$
$=3\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
Hence, Unit vector along $3\vec{\text{a}}-2\vec{\text{b}}=\frac{3\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}}{\sqrt{3^2+4^2+(-4)^2}}$
$=\frac{3\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}}{\sqrt{9+16+16}}$
$=\frac{1}{\sqrt{41}}\big(3\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}\big)$
View full question & answer→Question 862 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are non-collinear vectors, then find the value of $\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{i}}\big]\hat{\text{i}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{j}}\big]\hat{\text{j}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{k}}\big]\hat{\text{k}}.$
AnswerFor any vector $\vec{\text{r}},$ we have
$\big(\vec{\text{r}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{r}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{r}}.\hat{\text{k}}\big)\hat{\text{k}}=\vec{\text{r}}$
Replacing $\vec{\text{r}}$ by $\vec{\text{a}}\times\vec{\text{b}},$ we have
$\big[(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{i}}\big]+\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{j}}+\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{k}}\big]\hat{\text{k}}=\vec{\text{a}}\times\vec{\text{b}}$
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{i}}\big]\hat{\text{i}}\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{j}}\big]\hat{\text{j}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{k}}\big]\hat{\text{k}}=\vec{\text{a}}\times\vec{\text{b}}$
View full question & answer→Question 872 Marks
Write a unit vector perpendicular to $\hat{\text{i}}+\hat{\text{j}}$ and $\hat{\text{j}}+\hat{\text{k}}.$
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}};\vec{\text{b}}=0\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&0\\0&1&1 \end{vmatrix}$
$=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{1+1+1}$
$=\sqrt{3}$
Unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ is, $\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}=\frac{1}{\sqrt{3}}\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 882 Marks
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the given vectors are non-coplanar:
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}},\ \vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}$ and $\vec{\text{a}}+\vec{\text{b}}-3\vec{\text{c}}$
AnswerLet if possible the given vectors are coplanar. Then one of the vector is expressible in the terms of the other two.
We have,
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=\text{x}\big(\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}\big)+\text{y}\big(\vec{\text{a}}+\vec{\text{b}}-3\vec{\text{c}}\big)$
$=\vec{\text{a}}(\text{x + y})+\vec{\text{b}}(\text{x + y})+\vec{\text{c}}(-2\text{x}-3\text{y})$
$\Rightarrow\text{x + y}=2,\ \text{x + y}=-1,\ -2\text{x}-3\text{y}=3$
which is not true, as $\text{x + y}=2\neq-1$. Hence, the given vectors are non-coplanar.
View full question & answer→Question 892 Marks
Find a unit vector in the direction of $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$.
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{2^2+(-3)^2+6^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
Unit vector $=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}{7}$
$=\frac{2}7\hat{\text{i}}-\frac{3}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 902 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big).$
AnswerLet:
$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
$\vec{\text{b}}\times\vec{\text{a}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{a}_1&\text{a}_2&\text{a}_3\end{vmatrix}$
$=\hat{\text{i}}(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)-\hat{\text{j}}(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\hat{\text{k}}(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)$
Now,
$\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)$
$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big).\big[\hat{\text{i}}(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)\\-\hat{\text{j}}(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\hat{\text{k}}(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)\big]$
$=\text{a}_1(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)-\text{a}_2(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\text{a}_3(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)$
$=\text{a}_1\text{b}_2\text{a}_3-\text{a}_1\text{b}_3\text{a}_2-\text{a}_2\text{b}_1\text{a}_3+\text{a}_2\text{b}_3\text{a}_1+\text{a}_3\text{b}_1\text{a}_2-\text{a}_3\text{b}_2\text{a}_1$
$=0$
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If a vector makes angles $\alpha,\beta,\gamma$ with OX, OY and OZ respectively. then write the value of $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$.
AnswerSuppose, a vector $\overrightarrow{\text{OP}}$ makes an angle $\alpha,\beta,\gamma$ with OX, OY and OZ respectively.Then direction consines of the vector are given by $\text{l}=\cos\alpha,\ \text{m}=\cos\beta,\ \text{n}=\cos\gamma$Consider,
$\sin^2\alpha+\sin^2\beta+\sin^2\gamma\\=1-\cos^2\alpha+1-\cos^2\beta+1-\cos^2\gamma$
$=3-(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)$
$=3-(\text{l}^2+\text{m}^2+\text{n}^2)$
$=3-1$ $[\because\ \text{l}^2+\text{m}^2+\text{n}^2=1]$
$=2$
View full question & answer→Question 922 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 932 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 942 Marks
Write two different vectors having same direction.
AnswerLet $\vec{\text{p}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{q}}=2\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}$
Then, direction cosines of $\vec{\text{p}}$ are
$\text{l}=\frac{1}{\sqrt{1^2+2^2+3^2}}=\frac{1}{\sqrt{14}},\text{m}=\frac{2}{\sqrt{1^2+2^2+3^2}}=\frac{2}{\sqrt{14}}$ and $\text{n}=\frac{3}{\sqrt{1^2+2^2+3^2}}=\frac{3}{\sqrt{14}}$
Direction cosines of $\vec{\text{q}}$ are
$\text{l}=\frac{2}{\sqrt{2^2+4^2+6^2}}=\frac{2}{2\sqrt{14}}=\frac{1}{\sqrt{14}},$ $\text{m}=\frac{4}{\sqrt{2^2+4^2+6^2}}=\frac{4}{2\sqrt{14}}=\frac{2}{\sqrt{14}}$ and $\text{n}=\frac{6}{\sqrt{2^2+4^2+6^2}}=\frac{6}{2\sqrt{14}}=\frac{3}{\sqrt{14}}$
The direction cosines of two vectors are same. Hence the two different vectors $\vec{\text{p}},\vec{\text{q}}$ have same directions.
View full question & answer→Question 952 Marks
Show that the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ is equally inclined with the axes OX, OY and OZ.
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Then,
$|\vec{\text{a}}|=\sqrt{1^1+1^1+1^1}=\sqrt{3}$
Therefore, the direction ratios of $\vec{\text{a}}$ are $\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big).$
Now, let $\alpha,\beta$ and $\gamma$ be the angles format by $\vec{\text{a}}$ with the positive directions of x, y and z axes.
Then, we have $\cos\alpha=\frac{1}{\sqrt{3}},\cos\beta=\frac{1}{\sqrt{3}},\cos\gamma=\frac{1}{\sqrt{3}}.$
Hence, the given vector is equally inclined to axes OX, OY and OZ.
View full question & answer→Question 962 Marks
Write the direction cosines of the vector $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ .
AnswerGiven: $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Then, direction cosines are
$\frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}}$ or, $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
View full question & answer→Question 972 Marks
Find the components along the coordinate axis of the position vector of the following point:Q(-5, 1)
AnswerHere, Q = (-5, 1)
Position vector of $\text{Q}=-5\hat{\text{i}}+\hat{\text{j}}$
Component of Q along x-axis $=-5\hat{\text{i}}$
Component of Q along x-axis $=\hat{\text{j}}$
View full question & answer→Question 982 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 992 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 1002 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 $\frac{\pi}{6}.$
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