Question 12 Marks
Represent the following graphically:
- A displacement of 40km, 30º east of north.
- A displacement of 50km south-east.
- A displacement of 70km, 40º north of west.
Answer
- The vector $\overrightarrow{\text{OP}}$ represents the required displacement vector.
- The vector $\overrightarrow{\text{OQ}}$ represents the required vector.
- The vector $\overrightarrow{\text{OR}}$ represents the required vector.
View full question & answer→Question 22 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes:
$4\hat{\text{i}}+8\hat{\text{j}}+\hat{\text{k}}$
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 $\vec{\text{r}}=4\hat{\text{i}}+8\hat{\text{j}}+\hat{\text{k}}$ are proportional to 4, 8, 1. Therefore,
Direction cosine of $\vec{\text{r}}$ are $\frac{4}{\sqrt{4^2+8^2+1^2}},\frac{8}{\sqrt{4^2+8^2+1^2}},\frac{1}{\sqrt{4^2+8^2+1^2}}$ or $\frac{4}{9},\frac{8}{9},\frac{1}{9}.$
$\therefore\alpha=\cos^{-1}=\Big(\frac{4}{9}\Big),\beta=\cos^{-1}=\Big(\frac{8}{9}\Big),\gamma=\cos^{-1}=\Big(\frac{1}{9}\Big)$.
View full question & answer→Question 32 Marks
Find the sum of the following vectors: $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}},\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{c}}=2\hat{\text{i}}-3\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}},\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{c}}=2\hat{\text{i}}-3\hat{\text{k}}$So, Sum of the three vectors$=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{i}}+3\hat{\text{k}}$
$=5\hat{\text{i}}-5\hat{\text{j}}+3\hat{\text{k}}$
View full question & answer→Question 42 Marks
Find the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2).
AnswerThe position vector of mid-point R of the vector joining points P(2, 3, 4) and Q(4, 1, -2) is given by,
$\overrightarrow{\text{OR}}=\frac{\big(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}\big)+\big(4\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\big)}{2}$
$=\frac{(2+4)\hat{\text{i}}+(3+1)\hat{\text{j}}+(4-2)\hat{\text{k}}}{2}$
$=\frac{6\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}}{2}$
$=3\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 52 Marks
For what value of 'a' the vectors $2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$ are collinear?
AnswerGiven: Two vectors, let $\vec{\text{p}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{q}}=\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$
Since the given vectors are collinear, we have,
$\vec{\text{p}}=\lambda\vec{\text{q}}$
$\Rightarrow\ 2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}=\lambda\big(\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}\big)$
$\Rightarrow\ 2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}=\text{a}\lambda\hat{\text{i}}+6\lambda\hat{\text{j}}-8\lambda\hat{\text{k}}$
$\Rightarrow\ \lambda\text{a}=2,6\lambda=-3$ and $-8\lambda=4$
$\Rightarrow\ \lambda=-\frac{1}2$ and $\text{a}= -4$
View full question & answer→Question 62 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then write the values of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$.
AnswerLet, ABC be a given equilateral ttriangle and its vertices are $\text{A}(\vec{\text{a}}),\text{B}(\vec{\text{b}})$ and $\text{C}(\vec{\text{c}})$.
Also, $\text{O}(\vec{0})$ be the orthocentre of trianglre ABC.
We know that centroid and orthocentre of equilateral triangle coincide at one point.
Orthocentre of $\triangle\text{ABC}=\vec0$
$\Rightarrow$ Centroid $\triangle\text{ABC}=\vec0$
$\Rightarrow\ \frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3=\vec0$
$\therefore\ \vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec0$
View full question & answer→Question 72 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 82 Marks
If $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+4\hat{\text{j}}+9\hat{\text{k}}$, find a unit vector parallel to $\vec{\text{a}}+\vec{\text{b}}$.
AnswerGiven:$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}+4\hat{\text{j}}+9\hat{\text{k}}$
Now, $\vec{\text{a}}+\vec{\text{b}}=3\hat{\text{i}}+6\hat{\text{j}}+6\hat{\text{k}}$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{3^2+6^2+6^2}$
$=\sqrt{9+36+36}$ $=\sqrt{81}$ $=9$Unit vector parallel to $\vec{\text{a}}+\vec{\text{b}}=\frac{\vec{\text{a}}+\vec{\text{b}}}{\big|\vec{\text{a}}+\vec{\text{b}}\big|}=\frac{3\hat{\text{i}}+6\hat{\text{j}}+6\hat{\text{k}}}{9}$
$=\frac{1}9\times3\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)=\frac{1}3\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 92 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ represent the sides of a triangle taken in order, then write the value of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$.
AnswerLet ABC be a triangle such that $\overrightarrow{\text{BC}}=\vec{\text{a}},\ \overrightarrow{\text{CA}}=\vec{\text{b}},\ \overrightarrow{\text{AB}}=\vec{\text{c}}$. Then,$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}+\overrightarrow{\text{AB}}$
$=\overrightarrow{\text{BA}}+\overrightarrow{\text{AB}}$ $\Big[\because \overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\overrightarrow{\text{BA}}\Big]$
$=\vec0$
View full question & answer→Question 102 Marks
Find a vector of magnitude 4 units which is parallel to the vector $\sqrt3\hat{\text{i}}+\hat{\text{j}}$.
AnswerLet $\vec{\text{a}}=\sqrt3\hat{\text{i}}+\hat{\text{j}}$
Then, $\big|\vec{\text{a}}\big|=\sqrt{\big(\sqrt3\big)^2+1}=\sqrt{3+1}=\sqrt4=2$
A unit vector parallel to $\vec{\text{a}}=\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}2\big(\sqrt3\hat{\text{i}}+\hat{\text{j}}\big)$
Hence, Required vector $=4\hat{\text{a}}=4\times\frac{1}2\big(\sqrt3\hat{\text{i}}+\hat{\text{j}}\big)=2\sqrt3\hat{\text{i}}+2\hat{\text{j}}$
View full question & answer→Question 112 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 122 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of the vertices of a triangle, then write the position vector of its centroid.
AnswerLet ABC be a triangle and D, E and F are the midpoints of the sides BC, CA and AB respectively.
Also, Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of A, B, C respectively. Then the position vectors of D, E, F are $\Big(\frac{\vec{\text{b}}+\vec{\text{c}}}2\Big),\Big(\frac{\vec{\text{c}}+\vec{\text{a}}}2\Big),\Big(\frac{\vec{\text{a}}+\vec{\text{b}}}2\Big)$ respectively.
The position vector of a point divides AD in the ratio of 2; is $\frac{1.\vec{\text{a}}+2\frac{\vec{\text{b}}+\vec{\text{c}}}{2}}{2}=\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
Similarly, Position vectors of the points divides BE, CF in the ratio of 2 : 1 are equal to $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$.
Thus, the point dividing AD in the ratio 2 : 1 also divides BE, CF in the same ratio.
Hence, the medians of a triangle are concurrent and the position vector of the centroid is $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$.
View full question & answer→Question 132 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 142 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 152 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 162 Marks
If the position vector of a point (-4, -3) be $\vec{\text{a}}$, find $\big|\vec{\text{a}}\big|$.
AnswerGiven a point (-4, -3) such that its position vector $\vec{\text{a}}$ is given by
$\vec{\text{a}}=-4\hat{\text{i}}-3\hat{\text{j}}$
Then,
$\big|\vec{\text{a}}\big|=\sqrt{(-4)^2+(-3)^2}$
$=\sqrt{16+9}$
$=\sqrt{25}$
$=5$
View full question & answer→Question 172 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 $\big|3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}\big|$.
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}} $ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Now, $3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}=3\big(\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}\big)\\-2\big(\vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)+4\big(\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
$=9\hat{\text{i}}-3\hat{\text{j}}-12\hat{\text{k}}+4\hat{\text{i}}-8\hat{\text{j}}+6\hat{\text{k}}+4\hat{\text{i}}+8\hat{\text{j}}-4\hat{\text{k}}$
$=17\hat{\text{i}}-3\hat{\text{j}}-10\hat{\text{k}}$
Hence, $\big|3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}\big|$
$=\sqrt{17^2+(-3)^2+(-10)^2}$
$=\sqrt{289+9+100}$
$=\sqrt{398}$
View full question & answer→Question 182 Marks
Write two different vectors having same magnitude.
AnswerLet $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}=-2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
It can be observed that
$|\vec{\text{a}}|=\sqrt{2^2+(-1)^2+3^2}=\sqrt{14}$
$\big|\vec{\text{b}}\big|=\sqrt{(-2)^2+1^2+(-3)^2}=\sqrt{14}$
Hence, $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two vectors having same direction.
View full question & answer→Question 192 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 202 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 212 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 222 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 232 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 242 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 252 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 262 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 272 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 282 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 292 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 302 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 312 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 322 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 332 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 342 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 352 Marks
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 362 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 372 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 382 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 392 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 402 Marks
Find the position vector of the min-point of the line segment AB, where A is the point (3, 4, -2) and B is the point (1, 2, 4).
AnswerGiven: A(3, 4, -2) and B(1, 2, 4) Let C is the mid-point of AB $\therefore$ Position vector of $\text{C}=\frac{3\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}+\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}}2$$=\frac{4\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}}}2$
$=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 412 Marks
Write a unit vector making equal acute angles with the coordinates axes.
AnswerSuppose $\vec{\text{r}}$ makes an angle $\alpha$ wuth each of the axis OX, OY and OZ.
Then, its direction cosines are $\text{l}=\cos\alpha,\ \text{m}=\cos\alpha,\ \text{n}=\cos\alpha$.
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \text{l}^2+\text{l}^2+\text{l}^2=1$ $[\because\text{l = m = n}]$
$\Rightarrow\ 3\text{l}^2=1$
$\Rightarrow\ \text{l}^2=\frac{1}3$
$\Rightarrow\ \text{l}=\pm\frac{1}{\sqrt3}$
Since the angle is acute Hence, we take only positive value
Therefore, unit vector is $\Big(\frac{1}{\sqrt3}\hat{\text{i}}+\frac{1}{\sqrt3}\hat{\text{j}}+\frac{1}{\sqrt3}\hat{\text{k}}\Big)$.
View full question & answer→Question 422 Marks
Find the vector in the direction of vector $2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$ which has magnitude 21 units.
AnswerLet $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$
$\therefore\ |\vec{\text{a}}|=\sqrt{2^2+(-3)^2+6^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
Unit vector in the direction of $\vec{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}7$
$\therefore$ vector in the direction of vector $\vec{\text{a}}$ which has magnitude 21 units
$=21\times\Big(\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}7\Big)$
$=3\big(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}\big)$
$=6\hat{\text{i}}-9\hat{\text{j}}+18\hat{\text{k}}$
View full question & answer→Question 432 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 442 Marks
If $|\vec{\text{a}}|=4$ and $-3\leq\lambda\leq2$, then write the range of $|\lambda\vec{\text{a}}|$.
AnswerIt is given that
$-3\leq\lambda\leq2$
$\Rightarrow\ -3\times|\vec{\text{a}}|\leq\lambda|\vec{\text{a}}|\leq2\times|\vec{\text{a}}|$
$\Rightarrow\ -3\times4\leq|\lambda\vec{\text{a}}|\leq2\times4$ $\big(\text{k}|\vec{\text{a}}|=|\text{k}\vec{\text{a}}|,\ \text{k}$ is scalar$\big)$
$\Rightarrow\ -12\leq|\lambda\vec{\text{a}}|\leq8$
Thus, the range of $|\lambda\vec{\text{a}}|$ is [-12, 8]
View full question & answer→Question 452 Marks
Write the unit vector in the direction of $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$.
AnswerWe have,$\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{3^2+(-2)^2+6^2}$ $=\sqrt{9+4+36}$ $=\sqrt{49}$ $=7$ $\therefore$ Unit vector in the direction of $\vec{\text{a}}=\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}7\big(3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}\big)=\frac{3}7\hat{\text{i}}-\frac{2}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 462 Marks
If P, Q and R are three collinear points such that $\overrightarrow{\text{PQ}}=\vec{\text{a}}\text{ and }\overrightarrow{\text{QR}}=\vec{\text{b}}$. Find the vector $\overrightarrow{\text{PR}}$.
AnswerGiven that, P, Q, R are collinear.It also given that, $\overrightarrow{\text{PQ}}=\vec{\text{a}}\text{ and }\overrightarrow{\text{QR}}=\vec{\text{b}}$
$\overrightarrow{\text{PR}}=\overrightarrow{\text{PQ}}+\overrightarrow{\text{QR}}$
$=\vec{\text{a}}+\vec{\text{b}}$
$\overrightarrow{\text{PR}}=\vec{\text{a}}+\vec{\text{b}}$
View full question & answer→Question 472 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$, find a unit vector parallel to $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}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\therefore\ 2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)\\-\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)+3\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$
$=3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
A unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$ is given by $\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{(3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{3^2+(-3)^2+2^2}}$
$=\frac{(3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{22}}$
$=\frac{3}{\sqrt{22}}\hat{\text{i}}-\frac{3}{\sqrt{22}}\hat{\text{j}}+\frac{2}{\sqrt{22}}\hat{\text{k}}$
View full question & answer→Question 482 Marks
Write a vector of magnitude 12 units which makes 45º angle with x-axis, 60º angle with y-axis and an obtuse angle with z-axis.
AnswerSuppose a vector $\vec{\text{r}}$ makes an angle 45º with OX, 60º with OY and having magnitude 12 units. $\text{l}=\cos45^{\circ}=\frac{1}{\sqrt2}$ and $\text{m}=\cos60^{\circ}=\frac{1}2$Now, $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \frac{1}2+\frac{1}4+\text{n}^2=1$
$\Rightarrow\ \text{n}^2=\frac{1}4$
$\Rightarrow\ =-\frac{1}2$ $[\because$ The angle with the z-axis is obtuse$]$
Therefore,
$\vec{\text{r}}=|\vec{\text{r}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$
$=12\Big(\frac{1}{\sqrt2}\hat{\text{i}}+\frac{1}2\hat{\text{j}}-\frac{1}2\hat{\text{k}}\Big)$
$=6\big(\sqrt2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
View full question & answer→Question 492 Marks
If G denots the centroid of $\triangle\text{ABC}$, then write the value of $\overrightarrow{\text{GA}}+\overrightarrow{\text{GB}}+\overrightarrow{\text{GC}}$.
AnswerLet $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ be the position vectors of the vertices A, B, C respectively. Then, the position vector of the centroid G is $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
Thus,
$\overrightarrow{\text{GA}}+\overrightarrow{\text{GB}}+\overrightarrow{\text{GC}}$
$=\vec{\text{a}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)+\vec{\text{b}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)+\vec{\text{c}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-3\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\vec0$
View full question & answer→Question 502 Marks
Find the position vector of a point R which divides the line segment joining points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)$ and $\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1.Externally
AnswerGiven: R divides the line segment joining the points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big),\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1 externally.
Therefore position vector of $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=-3\hat{\text{i}}+\hat{\text{k}}$
View full question & answer→Question 512 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 522 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 532 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\vec{\text{a}}=-\vec{\text{b}}\Rightarrow\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$
AnswerTrue
$\vec{\text{a}}=-\vec{\text{b}}$
Take modulus both sides
$\big|\vec{\text{a}}\big|=\big|-\vec{\text{b}}\big|$
$\Rightarrow\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ $\Big[\therefore\ \big|-\vec{\text{b}}\big|=\big|\vec{\text{b}}\big|\Big]$
View full question & answer→Question 542 Marks
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ denote the position vectors of points A and B respectively and C is a point on AB such that 3AC = 2AB, then write the position vector of C.
AnswerGiven: $\vec{\text{a}}\text{ and }\vec{\text{b}}$ denote the position vectors of points A and B respectively and C is a point on AB such that 3AC = 2AB.Let $\vec{\text{c}}$ is the position vector of C.
Now, $\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$ $\overrightarrow{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}$ Consider, $3\text{AC}=2\text{AB}$ $\Rightarrow\ 3\big(\vec{\text{c}}-\vec{\text{a}}\big)=2\big(\vec{\text{b}}-\vec{\text{a}}\big)$ $\Rightarrow\ 3\vec{\text{c}}-3\vec{\text{a}}=2\vec{\text{b}}-2\vec{\text{a}}$ $\Rightarrow\ 3\vec{\text{c}}=2\vec{\text{b}}+\vec{\text{a}}$ $\Rightarrow\ \vec{\text{c}}=\frac{1}3\big(2\vec{\text{b}}+\vec{\text{a}}\big)$ $\Rightarrow\ \vec{\text{c}}=\frac{1}3\big(\vec{\text{a}}+2\vec{\text{b}}\big)$ Hence, the position vector of C is $\frac{1}3\big(\vec{\text{a}}+2\vec{\text{b}}\big)$
View full question & answer→Question 552 Marks
A vector $\vec{\text{r}}$ is inclined at equal acute angles to x-axis, y-axis and z-axis. If $|\vec{\text{r}}|=6$ units, find $\vec{\text{r}}$.
AnswerHere, $\alpha=\beta=\gamma$
$\Rightarrow\ \cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\ \text{l}=\text{m}=\text{n}=\text{x}$ (say)
We know that,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\text{x}^2+\text{x}^2+\text{x}^2=1$
$3\text{x}^2=1$
$\text{x}^2=\frac{1}3$
$\text{x}=\pm\frac{1}{\sqrt3}$
$\text{l}=\pm\frac{1}{\sqrt3},\ \text{m}=\pm\frac{1}{\sqrt3},\ \text{n}=\pm\frac{1}{\sqrt3}$
View full question & answer→Question 562 Marks
A unit vector $\vec{\text{r}}$ makes angles $\frac{\pi}3\text{ and }\frac{\pi}2$ with $\hat{\text{j}}\text{ and } \hat{\text{k}}$ respectively and an acute angle $\theta$ with $\hat{\text{i}}$. Find $\theta$.
AnswerA unit vector makes an angle $\frac{\pi}3\text{ and }\frac{\pi}2$ with $\hat{\text{j}}\text{ and } \hat{\text{k}}$Let l, m, n be its direction cosines
$\therefore\ \text{l}=\cos\theta,\ \text{m}=\cos\big(\frac{\pi}3\big)=\frac{1}2,\ \text{n}=\cos\big(\frac{\pi}2\big)=0$
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \text{l}^2+\frac{1}4+0=1$
$\Rightarrow\ \text{l}^2=1-\frac{1}4=\frac{3}4$
$\Rightarrow\ \text{l}=\pm\frac{\sqrt3}2$
$\therefore\ \vec{\text{r}}$ makes an acute angle 30º, 150º with $\hat{\text{i}}$
Since, angle $\theta$ is acute.
$\therefore\ \theta=30^{\circ}$
View full question & answer→Question 572 Marks
Write the position vector of a point dividing the line segment joining points having position vectors $\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ externally in the ratio 2 : 3.
AnswerLet A and B be the points with position vectors $\vec{\text{a}} = \hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}},\vec{\text{b}} = 2\hat{\text{i}} - \hat{\text{j}} + 3\hat{\text{k}} $ respectively.
Let C divide AB externally in the ratio 2 : 3 such that AC : CB = 2 : 3
$\therefore $ Postion vector of C $ = \frac{2\big(2\hat{\text{i}} - \hat{\text{j}} + 3\hat{\text{k}}\big) - 3 \big(\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}}\big)}{2 - 3}$
$ = \frac{4\hat{\text{i}} - 2\hat{\text{j}} + 6\hat{\text{k}} - 3\hat{\text{i}} - 3\hat{\text{j}} + 6\hat{\text{k}}}{-1}$
$= \frac{\hat{\text{i}} - 5\hat{\text{j}} + 12\hat{\text{k}}}{-1}$
$= -\hat{\text{i}} + 5\hat{\text{j}} - 12\hat{\text{k}}$
View full question & answer→Question 582 Marks
Write a unit vector in the direction of $\overrightarrow{\text{PQ}}$, where P and Q are the points (1, 3, 0) and (4, 5, 6) respectively.
AnswerP(1, 3, 0) and Q(4, 5, 6) are the given points.
$\therefore\ \overrightarrow{\text{PQ}}=\big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\big)-\big(\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}}\big)$
$=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\Rightarrow\ \Big|\overrightarrow{\text{PQ}}\Big|=\sqrt{3^2+2^2+6^2}$
$=\sqrt{9+4+36}$
$=\sqrt{49}$
$=7$
$\therefore$ Unit vector in the direction of $\overrightarrow{\text{PQ}}=\frac{\overrightarrow{\text{PQ}}}{\Big|\overrightarrow{\text{PQ}}\Big|}=\frac{3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}}7=\frac{1}7\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$
View full question & answer→Question 592 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\vec{\text{b}}$
AnswerFalse
$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\vec{\text{b}}$
Consider an example,
$\vec{\text{a}}=\text{i}+\sqrt3\text{j}$ and $\vec{\text{b}}=\sqrt2\text{i}+\sqrt2\text{j}$
$\big|\vec{\text{a}}\big|=\sqrt{1^2+\big(\sqrt3\big)^2}=2$ and $\big|\vec{\text{b}}\big|=\sqrt{\big(\sqrt2\big)^2+\big(\sqrt2\big)^2}=2$
Thus, $\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ but $\vec{\text{a}}\neq\vec{\text{b}}$
View full question & answer→Question 602 Marks
Find the position vector of a point R which divides the line segment joining points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)$ and $\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1.Internally
AnswerGiven: R divides the line segment joining the points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big),\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1 internally.
Therefore position vector of $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=\frac{1}3\big(-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}\big)$
View full question & answer→Question 612 Marks
Find a vector $\vec{\text{a}}$ of magnitude $5\sqrt2$, making an angle of $\frac{\pi}4$ with x-axis, $\frac{\pi}2$ with y-axis and an acute angle $\theta$ with z-axis.
AnswerIt is given that vector $\vec{\text{a}}$ makes an angle of $\frac{\pi}4$ with x-axis, $\frac{\pi}2$ with y-axis and an acute angle $\theta$ with z-axis.$\therefore\ \text{l}=\cos\frac{\pi}4=\frac{1}{\sqrt2},\text{m}=\cos\frac{\pi}2=0,\text{n}=\cos\theta$
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \frac{1}2+0+\cos^2\theta=1$
$\Rightarrow\ \cos^2\theta=1-\frac{1}2=\frac{1}2$
$\Rightarrow\ \cos\theta=\frac{1}{\sqrt2}$ ($\theta$ is acute)
We know that
$\vec{\text{a}}=|\vec{\text{a}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$
$\Rightarrow\ \vec{\text{a}}=5\sqrt2\Big(\frac{1}{\sqrt2}\hat{\text{i}}+0\hat{\text{j}}+\frac{1}{\sqrt2}\hat{\text{k}}\Big)$
$\Rightarrow\ \vec{\text{a}}=5\big(\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 622 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 632 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 642 Marks
Can a vector have direction angles 45º, 60º, 120º?
AnswerYes,
Let a vector makes an angle $\alpha=45^{\circ},\ \beta=60^{\circ},\ \gamma=120^{\circ}$ with OX, OY, OZ respectively.
Let l, m, n be the direction cosines of the vector. Then,
$\text{l}=\cos45^{\circ}=\frac{1}{\sqrt{2}}$
$\text{m}=\cos60^{\circ}=\frac{1}2$
$\text{n}=\cos120^{\circ}=-\frac{1}2$
So,
$\text{l}^2+\text{m}^2+\text{n}^2$
$=\frac{1}2+\frac{1}4+\frac{1}4$
$=1$
Since, the vector has direction cosines such that $\text{l}^2+\text{m}^2+\text{n}^2=1$
Hence, a vector can have direction angles 45º, 60º, 120º
View full question & answer→Question 652 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 662 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 672 Marks
Write a vector in the direction of vector $5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ which has magnitude of 8 unit.
AnswerGiven:
$\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{5^2+(-1)^2+2^2}$
$=\sqrt{25+1+4}$
$=\sqrt{30}$
$\therefore$ Position vector in the direction of vector $=8\times\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{8}{\sqrt{30}}\big(5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 682 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are position vectors of the points A, B and C respectively, write the value of$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{AC}}$.
AnswerGiven: $\vec{\text{a}},\vec{\text{b}}\text{ and }\vec{\text{c}}$ are the position vectors of A, B, C respectively. Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\overrightarrow{\text{BC}}=\vec{\text{c}}-\vec{\text{b}}$
$\overrightarrow{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}$
Therefore,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{AC}}\\=\vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}$
$=2\big(\vec{\text{c}}-\vec{\text{a}}\big)$
View full question & answer→Question 692 Marks
Find the unit vector in the direction of $3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}$.
AnswerLet $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}$Then, $\big|\vec{\text{a}}\big|=\sqrt{3^2+4^2+(-12)^2}$
$=\sqrt{9+16+144}$
$=\sqrt{169}$
$=13$
So, a unit vector in the direction of $\vec{\text{a}}$ is given by
$\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}{13}\big(3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}\big)$
$=\frac{3}{13}\hat{\text{i}}+\frac{4}{13}\hat{\text{j}}-\frac{12}{13}\hat{\text{k}}$
View full question & answer→Question 702 Marks
Write a unit vector in the direction of $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.
AnswerGiven:$\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\big|\vec{\text{b}}\big|=\sqrt{2^2+1^2+2^2}$ $=\sqrt{4+1+4}$ $=\sqrt9$ $=3$ $\therefore$ Unit vector $=\frac{\vec{\text{b}}}{\big|\vec{\text{b}}\big|}=\frac{1}3\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ $=\frac{2}3\hat{\text{i}}+\frac{1}3\hat{\text{j}}+\frac{2}3\hat{\text{k}}$
View full question & answer→Question 712 Marks
Prove that 1, 1, 1 cannot be direction cosines of a straight line.
AnswerLet 1, 1, 1 be the direction cosines of a straight line. Then,
$1^2+1^2+1^2=3\neq1$
Since direction cosines of a line which makes equal angle with the axes must satisfy
$\text{l}^2+\text{m}^2+\text{n}^2=1$
Hence 1, 1, 1 cannot be the direction cosines of a straight line.
View full question & answer→Question 722 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$, find the unit vector in the direction of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$.
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$
Then, $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{j}}+\hat{\text{k}}+\hat{\text{k}}+\hat{\text{i}}$
$=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$\therefore\ |\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}|=\sqrt{2^2+2^2+2^2}$
$=\sqrt{12}$
$=2\sqrt3$
Therefore, unit vector in the direction of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\frac{2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)}{2\sqrt3}=\frac{1}{\sqrt3}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 732 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes: $\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
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 $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ are proportional to 1, -1, 1. Therefore, Direction cosine of $\vec{\text{r}}$ are $\frac{1}{\sqrt{1^2+(-1)^2+1^2}},\frac{-1}{\sqrt{1^2+(-1)^2+1^2}},\frac{1}{\sqrt{1^2+(-1)^2+1^2}}$ or, $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$ $\therefore\cos\alpha=\frac{1}{\sqrt{3}},\cos\beta=\frac{1}{\sqrt{3}},\cos\gamma=\frac{1}{\sqrt{3}}$ $\alpha=\cos^{-1}=\Big(\frac{1}{\sqrt{3}}\Big),\beta=\cos^{-1}=\Big(\frac{-1}{\sqrt{3}}\Big),\gamma=\cos^{-1}=\Big(\frac{1}{\sqrt{3}}\Big)$
View full question & answer→Question 742 Marks
What is the cosine of the angle with the vector $\sqrt2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ makes with y-axis?
AnswerGiven $\sqrt2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Therefore, direction cosines are $\frac{\sqrt2}{\sqrt{(\sqrt2)^2+1^2+1^2}},\frac{1}{\sqrt{(\sqrt2)^2+1^2+1^2}},\frac{1}{\sqrt{(\sqrt2)^2+1^2+1^2}}$ or $\frac{1}{\sqrt2},\frac{1}2,\frac{1}2$
So, cosine angle with respect to y-axis is $\frac{1}2$
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