Question 1512 Marks
Write the direction cosines of the vector $\vec{\text{r}}=6\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$.
AnswerGiven: $\vec{\text{r}}=6\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Then, direction cosines of $\hat{\text{r}}$ are $\frac{6}{\sqrt{6^2+(-2)^2+3^2}},\frac{-2}{\sqrt{6^2+(-2)^2+3^2}},\frac{3}{\sqrt{6^2+(-2)^2+3^2}}$ or, $\frac{6}7,\frac{-2}7,\frac{3}7$
View full question & answer→Question 1522 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-coplanar vectors, then find the value of $\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$
AnswerWe have
$\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$
$\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}+\frac{\big[\vec{\text{b}}\vec{ \text{a}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}$ (By definition of scalar triple product)
$=\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}+\frac{-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}$ (Change of cyclic order of vectors changes the sign of the scalar triple product)
$=1-1$
$=0$
View full question & answer→Question 1532 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 1542 Marks
Find the components along the coordinate axis of the position vector of the following point:
P(3, 2)
AnswerHere, P = (3, 2)
Position vector of $\text{P}=3\hat{\text{i}}+2\hat{\text{j}}$
Component of P along x-axis $=3\hat{\text{i}}$
Component of P along x-axis $=2\hat{\text{j}}$
View full question & answer→Question 1552 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 1562 Marks
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the given vectors are non-coplanar:
$\vec{\text{a}}+2\vec{\text{b}}+3\vec{\text{c}},\ 2\vec{\text{a}}+\vec{\text{b}}+3\vec{\text{c}}$ and $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
AnswerLet if possible the given vectors are coplanar. Then one of the vector is expressible in the terms of the other two.
We have,
$\vec{\text{a}}+2\vec{\text{b}}+3\vec{\text{c}}=\text{x}\big(2\vec{\text{a}}+\vec{\text{b}}+3\vec{\text{c}}\big)+\text{y}\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\vec{\text{a}}(\text{2x + y})+\vec{\text{b}}(\text{x + y})+\vec{\text{c}}(3\text{x}+\text{y})$
$\Rightarrow\text{2x + y}=1,\ \text{x + y}=2,\ 3\text{x}+\text{y}=3$
On solving the first two equations we get x = -1, y = 3. Clearly the values of x, y does not satisfy the third equation.
Hence, the given vectors are non-coplanar.
View full question & answer→Question 1572 Marks
Write a unit vector in the direction of the sum of the vectors $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\text{y}\hat{\text{j}}-7\hat{\text{k}}$.
AnswerWe have, $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\text{y}\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\ \vec{\text{a}}+\vec{\text{b}}=\big(2\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}\big)+\big(2\hat{\text{i}}+\hat{\text{j}}-7\hat{\text{k}}\big)$
$=4\hat{\text{i}}+3\hat{\text{j}}-12\hat{\text{k}}$
$\Rightarrow\ \big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{4^2+3^2+(-12)^2}$
$=\sqrt{16+9+144}$
$=\sqrt{169}$
$=13$
$\therefore$ Required unit vector $=\frac{\vec{\text{a}}+\vec{\text{b}}}{\big|\vec{\text{a}}+\vec{\text{b}}\big|}=\frac{4\hat{\text{i}}+3\hat{\text{j}}-12\hat{\text{k}}}{13}$
$=\frac{4}{13}\hat{\text{i}}+\frac{3}{13}\hat{\text{j}}-\frac{12}{13}\hat{\text{k}}$
View full question & answer→Question 1582 Marks
Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes OX, OY and OZ.
Answer$\text{Let}\ \vec{a}=\hat{i}+\hat{j}+\hat{k}.$ Then,$|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$
Therefore, the direction cosines of $\vec{a}=\text{are}\ \bigg(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\bigg).$
Now, let $\alpha,\ \beta,$ and $\gamma$ be the angles formed by $\vec{a}$ with the positive directions of x, y, and z axes.
Then, we have $\text{cos}\ \alpha=\frac{1}{\sqrt{3}},\text{cos}\beta=\frac{1}{\sqrt{3}},\text{cos}\ \gamma=\frac{1}{\sqrt{3}}.$ Hence, the given vector is equally is equally inclined to axes OX, OY, and OZ.
View full question & answer→Question 1592 Marks
Find the components along the coordinate axis of the position vector of the following point:S(4,-3)
AnswerHere, S = (4, -3)
Position vector of $\text{S}=4\hat{\text{i}}-3\hat{\text{j}}$
Component of S along x-axis $=4\hat{\text{i}}$
Component of S along x-axis $=-3\hat{\text{j}}$
View full question & answer→Question 1602 Marks
If $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=144$ and $|\vec{\text{a}}|=4,$ find $\big|\vec{\text{b}}\big|.$
AnswerWe know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow144=4^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow144=16\big|\vec{\text{b}}\big|^2$
$\Rightarrow\big|\vec{\text{b}}\big|^2=9$
$\Rightarrow\big|\vec{\text{b}}\big|=3$
View full question & answer→Question 1612 Marks
Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
AnswerVertices of $\triangle\text{ABC}$ are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5). $\therefore\ \ \ \ \ \ \text{Position vector of point A}={(1, 1, 2) }=\hat{i}+\hat{j}+2\hat{k}$$\text{Position vector of point B}={(2, 3, 5) }=2\hat{i}+3\hat{j}+5\hat{k}$
$\text{Position vector of point C}={(1, 5, 5) }=\hat{i}+5\hat{j}+5\hat{k}$
$\text{Now}\ \ \overrightarrow{\text{AB}}\ \ $ = Position vector of point B - Position vector of point A$= 2\hat{i}+3\hat{j}+5\hat{k}-\big(\hat{i}+\hat{j}+2\hat{k}\big)$ $=2\hat{i}+3\hat{j}+5\hat{k}-\hat{i}-\hat{j}-2\hat{k}$
$=\hat{i}+2\hat{j}+3\hat{k}$
$\text{And}\ \ \overrightarrow{\text{AC}}\ \ $ = Position vector of point C - Position vector of point A$=\hat{i}+5\hat{j}+5\hat{k}\ -(\hat{i}+\hat{j}+2\hat{k})$ $=\hat{i}+5\hat{j}+5\hat{k}\ -\hat{i}-\hat{j}-2\hat{k}$
$=0\hat{i}+4\hat{j}+3\hat{k}$
$\therefore\ \ \ \ \ \ \overrightarrow{\text{AB}}\ \text{x}\ \overrightarrow{\text{AC}}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\0&4&3\end{vmatrix}=\hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0)=-6\hat{i}-3\hat{j}+4\hat{k}$ $\text{Now}\ \ \ \ \text{Area of triangle ABC}=\frac{1}{2}\bigg|\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}\bigg|$ $=\frac{1}{2}\sqrt{36+9+16}$$=\frac{1}{2}\sqrt{61}\ \text{sq.units}$
View full question & answer→Question 1622 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 1632 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 1642 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 1652 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 1662 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 1672 Marks
Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.
AnswerIf $\vec{\text{r}}$ is a unit vector in the XY-plane, then $\vec{\text{r}}=\cos\theta\hat{\text{i}}+\text{sin}\theta\hat{\text{j}}.$
Here, $\theta$ is the angle made by the unit vector with the positive direction of the x-axis.
Therefore, for $\theta$ = 30°:
$\vec{\text{r}}=\cos30^{\circ}\hat{\text{i}}+\text{sin}30^{\circ}\hat{\text{j}}=\frac{\sqrt{3}}{2}\hat{\text{i}}+\frac{1}{2}\hat{\text{j}}$
Hence, the required unit vector is $\frac{\sqrt{3}}{2}\hat{\text{i}}+\frac{1}{2}\hat{\text{j}}$
View full question & answer→Question 1682 Marks
Find the unit vector in the direction of vector $\overrightarrow{\text{PQ}}$ where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
AnswerThe given points are P (1, 2, 3) and Q (4, 5, 6).$\therefore\overrightarrow{\text{PQ}}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}$
$\Bigg|\overrightarrow{\text{PQ}}\Bigg|=\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=\sqrt{27}=3\sqrt{3} $
Hence, the unit vector in the direction of $\overrightarrow{\text{PQ}}$ is $\frac{\overrightarrow{\text{PQ}}}{\Big|\overrightarrow{\text{PQ}}\Big|}=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}$
View full question & answer→Question 1692 Marks
Prove that $\big(\vec{\text{a}}+\vec{\text{b}}\big)\cdot\big(\vec{\text{a}}+\vec{\text{b}}\big)=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2,$ if and only if $\vec{\text{a}},\vec{\text{b}}$ are perpendicular, given $\vec{\text{a}}\neq\vec{\text{0}},\vec{\text{b}}\neq\vec{\text{0}}.$
Answer$\Big(\vec{\text{a}}+\vec{\text{b}}\Big)\cdot\Big(\vec{\text{a}}+\vec{\text{b}}\Big)=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2$
$\Leftrightarrow\vec{\text{a}}\cdot\vec{\text{a}}+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}\cdot\vec{\text{b}}=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2$ [Distributivity of scalar products over addition]
$\Leftrightarrow\big|\vec{\text{a}}\big|^2+2\vec{\text{a}}\cdot\vec{\text{b}}+\Big|\vec{\text{b}}\Big|^2=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2\ \ \ $ $\Big[\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}}\ \text{(Scalar product is commutative)}\Big]$
$\Leftrightarrow2\vec{\text{a}}\cdot\vec{\text{b}}=0$
$\Leftrightarrow\vec{\text{a}}\cdot\vec{\text{b}}=0$
$\therefore\vec{\text{a}}\ \text{and}\ \vec{\text{b}} \ \text{are perpendicular.}$ $\ \ \ \Big[\vec{\text{a}}\neq\vec{\text{0}},\ \vec{\text{b}}\neq\vec{\text{0}}\ \text{(Given)}\Big]$
View full question & answer→Question 1702 Marks
Find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big), $ if $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.
AnswerThe given vectors are $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\vec{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Now,
$\vec{\text{b}}\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-1&2&1\\3&1&2 \end{vmatrix}=3\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)=\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big).\big(3\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}\big)$
$=2\times3+1\times5=3\times(-7)$
$=6+5-21=-10$
View full question & answer→Question 1712 Marks
$\text{Find}\ |\vec{a}\ \times\vec{b}|,\ \text{if}\ \vec{a}=\hat{i}-7\hat{j}+7\hat{k}\ \text{and}\ \vec{b}=3\hat{i}-2\hat{j}+2\hat{k}.$
Answer$\text{Give:}\ \ \vec{a}=\hat{i}-7\hat{j}+7\hat{k}\ \text{and}\ \vec{b}=3\hat{i}-2\hat{j}+2\hat{k}$$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-7&7\\3&-2&2\end{vmatrix}$
Expanding along first row, $\vec{a}\times\vec{b}=\hat{i}\begin{vmatrix}-7&7\\-2&2\end{vmatrix}-\hat{j}\begin{vmatrix}1&7\\3&2\end{vmatrix}+\hat{k}\begin{vmatrix}1&-7\\3&-2\end{vmatrix}$ $=\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)$$=0\hat{i}+19\hat{j}+19\hat{k}$
$\therefore\ \ \Big|\vec{a}\times\vec{b}\Big|=\sqrt{(0)^2+(19)^2+(19)^2}=\sqrt{2(19)^2}=19\sqrt{2}$
View full question & answer→Question 1722 Marks
Let the vectors $\vec{a},\vec{b},\vec{c}$ be given as $a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k},\ b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k},$ $c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}.$ Then show that $\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}.$
AnswerGiven: $\text{Vector}\ \vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k},\ \vec{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}$ $\text{and}\ \vec{c}=c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}$ $\therefore\ \ \vec{b}+\vec{c}=(b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}$ $\text{Now}\ \ \ \ \text{L.H.S}=\vec{a}\times\big(\vec{b}+\vec{c}\big)= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ b_1+c_1& b_2+c_2 & b_3+c_3 \end{vmatrix}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ b_1& b_2& b_3\end{vmatrix}+\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ c_1& c_2& c_3\end{vmatrix}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[By Property of Determinants]}$$=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}=\text{R.H.S}.$
View full question & answer→Question 1732 Marks
Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2).$
AnswerThe vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ can be obtained by,
$\overrightarrow{\text{PQ}}$ = Position vector of $Q -$ Position vector of $P$
$=(\text{x}_{2}-\text{x}_{1})\hat{\text{i}}+(\text{y}_{2}-\text{y}_{1})\hat{\text{j}}+(\text{z}_{2}-\text{z}_{1})\hat{\text{k}}$
$\bigg|\overrightarrow{\text{PQ}}\bigg|=\sqrt{(\text{x}_{2}-\text{x}_{1})^{2}+(\text{y}_{2}-\text{y}_{1})^{2}+(\text{z}_{2}-\text{z}_{1})^{2}}$
Hence, the scalar components and the magnitude of the vector joining the given points are respectively $ \{(\text{x}_{2}-\text{x}_{1}),(\text{y}_{2}-\text{y}_{1}),(\text{z}_{2}-\text{z}_{1})\}\ \text{and}$ $\sqrt{(\text{x}_{2}-\text{x}_{1})^{2}+(\text{y}_{2}-\text{y}_{1})^{2}+(\text{z}_{2}-\text{z}_{1})^{2}}$
View full question & answer→Question 1742 Marks
Show that the following triads of vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+7\hat{\text{k}},\vec{\text{c}}=5\hat{\text{i}}+6\hat{\text{j}}+5\hat{\text{k}}$
AnswerWe know that three vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff their scalar triple product is zero i.e. $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}1&2&-1\\3&2&7\\5&6&5 \end{vmatrix}$
$=1(1-42)-2(15-35)-1(18-10)$
$=-60+126-66$
$=0$
Hence, the given vectors are coplanar.
View full question & answer→Question 1752 Marks
Evaluate the following:$\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]$
AnswerWe have,
$\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat {\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]\\=(2\hat{\text{i}}\times\hat{\text{j}}).\hat{\text{k}}+(\hat{\text{i}}\times\hat{\text{k}}).2\hat{\text{i}}$
$=2\hat{\text{k}}.\hat{\text{k}}+(-\hat{\text{j}}).\hat{\text{j}}+(-\hat{\text{i}}).2\hat{\text{i}}$
$=2-1-2$
$=-1$
Therefore, $\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]=-1$
View full question & answer→Question 1762 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 1772 Marks
Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}\ \text{and}\ \vec{a}-\vec{b},\ \text{where}$ $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}-2\hat{k}.$
Answer$\text{Given:}\ \ \vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}-2\hat{k}$
$\text{On Adding}\ \vec{c}=\vec{a}+\vec{b}=3\hat{i}+2\hat{j+2\hat{k}}\ +\ \hat{i}+2\hat{j}-2\hat{k}$ $=4\hat{i}+4\hat{j}+0\hat{k}$
$\text{On Subtracting}\ \ \ \vec{d}=\vec{a}-\vec{b}=3\hat{i}+2\hat{j}+2\hat{k}\ - \ \hat{i}-2\hat{j}+2\hat{k}$ $=2\hat{i}+0\hat{j}+4\hat{k}$
$\text{Therefore,}\ \ \ \vec{n}=\vec{c}\times\vec{d}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\4&4&0\\2&0&4 \end{vmatrix}$
Expanding along first row $=\hat{i}(16-0)-\hat{i}(16-0)+\hat{k}(0-8)$
$\Rightarrow\ \ \vec{n}=16\hat{i}-16\hat{j}-8\hat{k}$
$\therefore\ \big|\vec{n}\big|=\sqrt{(16)^2+(-16)^2+(-8)^2}$ $\sqrt{256+256+64}=\sqrt{576}=24$
Therefore, a unit vector perpendicular to both $\vec{a}\ \text{and}\ \vec{b}\ \text{is}$
$\hat{n}=\pm\frac{\vec{n}}{|\vec{n}|}=\pm\frac{(16\hat{i}-16\hat{j}-8\hat{k})}{24}$
$=\pm\Big(\frac{16}{24}\hat{i}-\frac{16}{24}\hat{j}-\frac{8}{24}\hat{k}\Big)=\pm\Big(\frac{2}{3}\hat{i}-\frac{2}{3} \hat{j}-\frac{1}{3}\hat{k}\Big)$
View full question & answer→Question 1782 Marks
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).
AnswerThe vector with the initial point P (2, 1) and terminal point Q (-5, 7) can be given by, $\overrightarrow{\text{PQ}}=(-5-2)\hat{i}+(7-1)\hat{j}$ $\Rightarrow\ \overrightarrow{\text{PQ}}=-7\hat{i}+6\hat{j}$ Hence, the required scalar components are -7 and 6 while the vector component are, $-7\hat{i}\ \text{and}\ 6\hat{j}$
View full question & answer→Question 1792 Marks
$\text{Given that}\ \vec{a}\cdot\vec{b}=0\ \text{and}\ \vec{a}\times\vec{b}=\vec{0}.$ What can you conclude about the vectors $\vec{a}\ \text{and}\ \vec{b}$?
Answer$\text{Given:}\ \ \vec{a}.\vec{b}=0\ \Rightarrow\ \ \big|\vec{a}\big|.\big|\vec{b}\big|\cos\theta=0$ $\therefore\ \ \big|\vec{a}\big|=0\ \ \text{or} \ \big|\vec{b}\big|=0$$\ \text{or}\ \cos\theta=0\ \ \Rightarrow\ \ \theta=90^\circ$ $\Rightarrow \ \ \vec{a}=0\ \ \text{or}\ \ \vec{b}=0\ \ \text{or}$ $\ \ \text{vector}\ \vec{a}\ \text{is perpendicular to}\ \vec{b}.\ \ \ \ ......\text{(i)}$ $\text{Again, given}\ \vec{a}\times\vec{b}=0\ \Rightarrow\ \big|\vec{a}\times\vec{b}\big|=0$ $\ \Rightarrow\ \ \big|\vec{a}\big|.\big|\vec{b}\big|\sin\theta=0$ $\therefore\ \ \big|\vec{a}\big|=0\ \ \text{or}\ \ \big|\vec{b}\big|=0\ \ \text{or}\ \ $ $\sin\theta=0\ \ \Rightarrow\ \theta=0^\circ$ $\Rightarrow\ \ \vec{a}=0\ \ \text{or}\ \ \vec{b}=0 \ \ \text{or}\ \ $ $\text{vector}\ \vec{a}\ \text{and}\ \vec{b}\ \text{are collinear or parallel.}\ \ \ \ \ ...\text{(ii)}$ Since, vectors $\vec{a}\ \&\ \vec{b}$ are perpendicular to each other as well as parallel are not possible. ...(iii) Therefore, form eq. (i), (ii) and (iii), $\ \text{either}\ \vec{a}=\vec{0}\ \ \ \text{or}\ \vec{b}=\vec{0}$ $\therefore\ \ \vec{a}.\vec{b}=0 \ \ \text{and}\ \ \vec{a}\times\vec{b}=0$
View full question & answer→Question 1802 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}},\ \vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$, find $|3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}|$.
AnswerGiven $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}},\ \vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Now, $3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}=3\big(3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}\big)-2\big(-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)+4\big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
$=9\hat{\text{i}}-3\hat{\text{j}}-12\hat{\text{k}}+4\hat{\text{i}}-8\hat{\text{j}}+6\hat{\text{k}}+4\hat{\text{i}}+8\hat{\text{j}}-4\hat{\text{k}}$
$=17\hat{\text{i}}-3\hat{\text{j}}-10\hat{\text{k}}$
$\therefore\ |3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}|=\sqrt{17^2+(-3)^2+(-10)^2}$
$=\sqrt{289+9+100}$
$=\sqrt{398}$
View full question & answer→Question 1812 Marks
If $\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$ are two equal vectors, then write the value of x + y + z.
AnswerGiven: $\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$ Since the two vectors are equal. We have,$\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}=\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\ \text{x}=3,\ \text{y}=-2,\ \text{z}=-1$
$\therefore\ \text{x + y + z}=3-2-1=0$
View full question & answer→Question 1822 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 1832 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 1842 Marks
Write the position vector of a point dividing the line segment joining points A and B with position vectors $\vec{\text{a}}\text{ and }\vec{\text{b}}$ externally in the ratio 1 : 4, where $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$.
AnswerThe position vectors of A and B are
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Let C divides AB in the ratio such that AB : CB = 1 : 4
Position vector of $\text{C}=\frac{1\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)-4\big(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}\big)}{1-4}$
$=\frac{-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}-8\hat{\text{i}}-12\hat{\text{j}}-16\hat{\text{k}}}{-3}$
$=\frac{-9\hat{\text{i}}-11\hat{\text{j}}-15\hat{\text{k}}}{-3}$
$=3\hat{\text{i}}+\frac{11\hat{\text{j}}}3+5\hat{\text{k}}$
View full question & answer→Question 1852 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 1862 Marks
Find the unit vector in the direction of the vector $\vec{a}=\hat{i}+\hat{j}+2\hat{k}.$
AnswerThe unit vector $\hat{a}$ in the direction of vector $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ is given by $\hat{a}=\frac{\vec{a}}{|a|}.$
$\Big|\vec{a}\Big|=\sqrt{1^2+1^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$
$\therefore{\hat{a}}=\frac{\vec{a}}{\big|\vec{a}\big|}=\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}$
View full question & answer→Question 1872 Marks
For given vectors, $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}\ \ \text{and}\ \vec{b}=-\hat{i}+\hat{j}-\hat{k},$ find the unit vector in the direction of the vector $\vec{a}+\vec{b}.$
AnswerThe given vectors are $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}\ \ \text{and}\ \vec{b}=-\hat{i}+\hat{j}-\hat{k}$ $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$ $\therefore\vec{a}+\vec{b}=(2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}$ $=1\hat{i}+0\hat{j}+1\hat{k}=\hat{i}+\hat{k}$ $\big|\vec{a}+\vec{b}\big|=\sqrt{1^2+1^2}=\sqrt{2}$Hence, the unit vector in the direction of $\big(\vec{a}+\vec{b}\big)$ is
$\frac{\big(\vec{a}+\vec{b}\big)}{\big|\vec{a}+\vec{b}\big|}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$
View full question & answer→Question 1882 Marks
Find a unit vector in the direction of the vector $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}$ Then, $|\vec{\text{a}}|=\sqrt{3^2+(-2)^2+6^2}$ $=\sqrt{9+4+36}$ $=\sqrt{49}$ $=7$ $\therefore$ Unit vector $=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}}7$$=\frac{3}7\hat{\text{i}}-\frac{2}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 1892 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 1902 Marks
Write the length (magnitude) of a vector whose projections on the coordinate axes are 12, 3 and 4 units.
AnswerGiven: Projections on the coordinate axes are 12, 3, 4 units. Therefore,
Length of vector $=\sqrt{12^2+3^2+4^2}$
$=\sqrt{169}$
$=13$
View full question & answer→Question 1912 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 1922 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 1932 Marks
If $\overrightarrow{\text{AO}}+\overrightarrow{\text{OB}}=\overrightarrow{\text{BO}}+\overrightarrow{\text{OC}}$, prove that A, B, C are collinear points.
AnswerHere, $\overrightarrow{\text{AO}}+\overrightarrow{\text{OB}}=\overrightarrow{\text{BO}}+\overrightarrow{\text{OC}}$
$\overrightarrow{\text{OA}}-\overrightarrow{\text{BO}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{CO}}$
$\overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$ but $\vec{\text{B}}$ is a common vector. Hence,
A, B, C are collinear.
View full question & answer→Question 1942 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes:
$\hat{\text{j}}-\hat{\text{k}}$
AnswerLet $\vec{\text{r}}$ be the given vector, and let it make an angle $\alpha,\beta,\gamma$ with OX, OY, OZ respectively. Then, its direction cosines are $\cos\alpha,\cos\beta,\cos\gamma$.So direction ratios of $\vec{\text{r}}=\hat{\text{j}}-\hat{\text{k}}$ are proportional to 0, 1, -1. Therefore,
Direction cosine of $\vec{\text{r}}$ are $\frac{0}{\sqrt{0+1^2+(-1^2)}},\frac{1}{\sqrt{0+1^2+(-1^2)}},\frac{-1}{\sqrt{0+1^2+(-1^2)}}$ or $0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}$
$\therefore\cos\alpha=0,\cos\beta=\frac{1}{\sqrt{2}},\cos\gamma=\frac{-1}{\sqrt{2}}$
$\Rightarrow\alpha=\cos^{-1}=(0),\beta=\cos^{-1}=\Big(\frac{-1}{\sqrt{2}}\Big),\gamma=\cos^{-1}=\Big(\frac{-1}{\sqrt{2}}\Big)$
$\Rightarrow\alpha=\frac{\pi}{2},\beta=\frac{\pi}{4},\gamma=\frac{3\pi}{4}$
View full question & answer→Question 1952 Marks
Find the direction cosines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1), directed from A to B.
AnswerThe given points are A (1, 2 -3) and B (-1, -2, 1).
$\therefore{\overrightarrow{\text{AB}}}=(-1-1)\hat{i}+(-2-2)\hat{j}+\{1-(-3)\}\hat{k}$
$\Rightarrow\overrightarrow{\text{AB}}=-2\hat{i}-4\hat{j}+4\hat{k}$
$\therefore\bigg|\overrightarrow{\text{AB}}\bigg|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6$
Hence, the direction cosines of $\overrightarrow{\text{AB}}\ \text{are}\ \bigg(-\frac{2}{6},-\frac{4}{6},\frac{4}{6}\bigg)=\bigg(-\frac{1}{3},\frac{2}{3},\frac{2}{3}\bigg).$
View full question & answer→Question 1962 Marks
Find the values of x and y so that the vectors $2\hat{i}+3\hat{j}\ \text{and}\ x\hat{i}\ +y\hat{j}$ are equal.
AnswerThe two vectors $2\hat{i}+3\hat{j}\ \text{and}\ x\hat{i}\ +y\hat{j}$ will be equal if their corresponding components are equal.
Hence, the required values of x and y are 2 and 3 respectively.
View full question & answer→Question 1972 Marks
$\text{Find}\ \big|\vec{x}\big|,$ if for a unit unit vector $\vec{a},\ (\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=12.$
Answer$\text{Given:}\ \vec{a}\ \text{is a unit vector}\ \Rightarrow\ \big|\vec{a}\big|=1\ \ .......(\text{i})$
$\big(\vec{x}-\vec{a}\big)\big(\vec{a}+\vec{a}\big)=12\ $ $\Rightarrow\ \vec{x}.\vec{x}+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\vec{a}.\vec{a}=12$
$\Rightarrow\ \ \big|\vec{x}\big|^2+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\big|\vec{a}\big|^2=12\ $ $\Rightarrow\ \big|\vec{x}\big|^2+\vec{a}.\vec{x}-\vec{a}.\vec{x}-\big|\vec{a}\big|^2=12$
$ \Rightarrow\ \ \big|\vec{x}\big|^2-\big|\vec{a}\big|^2=12$
$\text{Putting}\ \big|\vec{a}\big|=1\ \text{from eq. (i),}\ \ \ \ \big|\vec{x}\big|^2-1=12$
$\Rightarrow\ \big|\vec{x}\big|^2=13\ \ \Rightarrow\ \ \big|\vec{x}\big|=13$
View full question & answer→Question 1982 Marks
Evaluate the following:
$\big[\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{j}}\hat{\text{k}}\hat{\text{i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]$
AnswerWe have,
$\big[\hat{\text{i}}\hat{\text{ j}}\hat{\text{ k}}\big]+\big[\hat{\text{j}}\hat{\text{ k}}\hat{\text{ i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]$
$=(\hat{\text{i}}\times\hat{\text{j}}).\hat{\text{k}}(\hat{\text{j}}\times\hat{\text{k}}).\hat{\text{i}}+(\hat{\text{k}}\times\hat{\text{i}}).\hat{\text{j}}$
$=\hat{\text{k}}.\hat{\text{k}}+\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}$
$=1+1+1$
$=3$
Therefore, $\big[\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{j}}\hat{\text{k}}\hat{\text{i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]=3$
View full question & answer→Question 1992 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 2002 Marks
If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}\ \text{with}\ \hat{i},\frac{\pi}{4}\ \text{with}\ \hat{j}$ and an acute angle $\theta$ with $\hat{k},$ then find $\theta$ and hence, the components of $\vec{a}.$
Answer$\text{Let}\ \ \ \hat{a}=\text{x}\hat{i}+\text{y }\hat{j}+\text{z}\hat{k}\ \text{be a unit vector.}\ \ \ \ ......\text{(i)}$ $\Rightarrow\ \ |\hat{a}|=1\ \ \Rightarrow\ \ \sqrt{\text{x}^2+\text{y}^2+\text{z}^2}=1$ $\text{Squaring both sides,}\ \ \ \ \ \text{x}^2+\text{y}^2+\text{z}^2=1\ \ \ .....\text{(ii)}$ $\text{Given:}\ \ \text{Angle between vectors}\ \hat{a}\ \text{and}\ \hat{i}=\hat{i}+0\hat{j}+0\hat{k}\ \text{is}\ \frac{\pi}{3}.$ $\therefore\ \cos\frac{\pi}{3}=\frac{\hat{a}.\hat{i}}{\big|\hat{a}\big|.\big|\hat{i}\big|}\ $ $\Rightarrow\ \ \frac{1}{2}=\frac{\text{x}(1)+\text{y}(0)+\text{z}(0)}{(1)(1)}$ $\Rightarrow\ \ \ \frac{1}{2}=\text{x}\ \ \ ....\text{(iii)}$Again, given $\text{Angel between vectors}\ \hat{a}\ \text{and}\ \hat{j}=0\hat{i}+\hat{j}+0\hat{k}\ \text{is}\ \frac{\pi}{4}.$
$\therefore\ \cos\frac{\pi}{4}=\frac{\hat{a}.{\hat{j}}}{|\hat{a}|.|\hat{j}|}\ \Rightarrow\ \frac{1}{\sqrt{2}}=\frac{\text{x}(0)+\text{y}(1)+\text{z}(0)}{(1)(1)}$
$\Rightarrow\ \frac{1}{\sqrt{2}}=\text{y}\ \ ......\text{(iv)}$
Again, given $\text{Angel between vectors}\ \hat{a}\ \text{and}\ \hat{k}=0\hat{i}+0\hat{j}+\hat{k}\ \text{is}\ \theta,\ \text{where}\ \theta\ \text{is acute angle.}$
$\therefore\ \cos\theta=\frac{\hat{a}.{\hat{k}}}{|\hat{a}|.|\hat{k}|}\ $ $\Rightarrow\ \ \cos\theta=\frac{\text{x}(0)+\text{y}(0)+\text{z}(1)}{(1)(1)}$ $\Rightarrow\ \ \cos\theta=\text{z}\ \ \ .......\text{(v)}$ Putting the values of x, y and z in eq. (ii),$\frac{1}{4}+\frac{1}{2}+\cos^2\theta=1$ $\ \Rightarrow\ \ \cos^2\theta=1-\frac{1}{4}-\frac{1}{2}$
$\Rightarrow\ \ \ \ \ \ \ \cos^2\theta=\frac{4-1-2}{4}=\frac{1}{4}$ $\ \Rightarrow\ \ \cos\theta=\pm\frac{1}{2}$ Since $\theta$ is acute angle, therefore cos $\theta$ is positive and hence $\frac{1}{2}=\cos\frac{\pi}{3}\ \Rightarrow\ \ \theta=\frac{\pi}{3}$ From eq. (v), $\ \ \text{z}=\cos\theta=\frac{1}{2}$ Putting values of x, y and z in eq. (i), $\ \ \hat{a}=\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k}$$\therefore\ \ \text{Components of}\ \hat{a}\ \text{are coefficients of}\ \hat{i},\hat{j},\hat{k}\ \text{in}\ \hat{a}$
$\Rightarrow\ \ \frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\ \text{and angle}\ \theta=\frac{\pi}{3}$
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