MCQ 2011 Mark
If the direction cosines of a line are $\left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right)$, then
- A
- B
$a>2$
- C
$a>0$
- ✓
$a= \pm \sqrt{3}$
AnswerCorrect option: D. $a= \pm \sqrt{3}$
(d) : Given that the direction cosines of a line are
$
\left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right) \text {. }
$
We know that the sum of squares of the direction cosines is 1 .
$
\Rightarrow \frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{a^2}=1 \Rightarrow \frac{3}{a^2}=1 \Rightarrow a^2=3 \Rightarrow a= \pm \sqrt{3}
$
View full question & answer→MCQ 2021 Mark
The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is
- ✓
$5 \hat{i}-4 \hat{j}+6 \hat{k}+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$
- B
$3 \hat{i}+7 \hat{j}+2 \hat{k}+\lambda(5 \hat{i}+4 \hat{j}+6 \hat{k})$
- C
$-5 \hat{i}+3 \hat{j}+4 \hat{k}+\lambda(7 \hat{i}+6 \hat{j}+12 \hat{k})$
- D
AnswerCorrect option: A. $5 \hat{i}-4 \hat{j}+6 \hat{k}+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$
(a): We have, $\vec{a}=5 \hat{i}-4 \hat{j}+6 \hat{k}$ and $\vec{b}=3 \hat{i}+7 \hat{j}+2 \hat{k}$
Therefore, the vector equation will be $\vec{r}=\vec{a}+\lambda \vec{b}$
$
\Rightarrow \vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})
$
View full question & answer→MCQ 2031 Mark
The value of $p$, so that the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}$ $=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ intersect at right angle, is
- A
$\frac{10}{11}$
- ✓
$\frac{70}{11}$
- C
$\frac{10}{7}$
- D
$\frac{70}{9}$
AnswerCorrect option: B. $\frac{70}{11}$
(b) : Equation of the given lines can be written in the standard form as
$
\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2} \text { and } \frac{x-1}{-\frac{3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}
$
$\because \quad$ Lines are perpendicular to each other.
$\therefore \quad a_1 a_2+b_1 b_2+c_1 c_2=0$
$
\Rightarrow(-3)\left(\frac{-3 p}{7}\right)+\left(\frac{2 p}{7}\right)(1)+(2)(-5)=0 \Rightarrow p=\frac{70}{11}
$
View full question & answer→MCQ 2041 Mark
If the line joining $(2,3,-1)$ and $(3,5,-3)$ is perpendicular to the line joining $(1,2,3)$ and $(3,5, \lambda)$, then $\lambda=$
Answer(d) : DR's of the given lines are 1, 2, -2 and 2, 3, $\lambda-3$.
Since, lines are perpendicular. $\therefore a_1 a_2+b_1 b_2+c_1 c_2=0$
$
\Rightarrow \quad 1 \times 2+2 \times 3-2(\lambda-3)=0 \Rightarrow \lambda=7
$
View full question & answer→MCQ 2051 Mark
The vector equation of the line through the points $A(3,4,-7)$ and $B(1,-1,6)$ is
- A
$\vec{r}=(3 \hat{i}-4 \hat{j}-7 \hat{k})+\lambda(\hat{i}-\hat{j}+6 \hat{k})$
- B
$\vec{r}=(\hat{i}-\hat{j}+6 \hat{k})+\lambda(3 \hat{i}-4 \hat{j}-7 \hat{k})$
- ✓
$\vec{r}=(3 \hat{i}+4 \hat{j}-7 \hat{k})+\lambda(-2 \hat{i}-5 \hat{j}+13 \hat{k})$
- D
$\vec{r}=(\hat{i}-\hat{j}+6 \hat{k})+\lambda(4 \hat{i}+3 \hat{j}-\hat{k})$
AnswerCorrect option: C. $\vec{r}=(3 \hat{i}+4 \hat{j}-7 \hat{k})+\lambda(-2 \hat{i}-5 \hat{j}+13 \hat{k})$
(c): If $\vec{a}=P . V$. of $A=3 \hat{i}+4 \hat{j}-7 \hat{k}$
and $\vec{b}=$ P.V. of $B=\hat{i}-\hat{j}+6 \hat{k}$, then the equation of line $A B$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
$
\therefore \quad \vec{r}=(3 \hat{i}+4 \hat{j}-7 \hat{k})+\lambda(-2 \hat{i}-5 \hat{j}+13 \hat{k})
$
View full question & answer→MCQ 2061 Mark
The equation of the line joining the points $(-3,4,11)$ and $(1,-2,7)$ is
- A
$\frac{x+3}{2}=\frac{y-4}{3}=\frac{z-11}{4}$
- ✓
$\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}$
- C
$\frac{x+3}{-2}=\frac{y+4}{3}=\frac{z+11}{4}$
- D
$\frac{x+3}{2}=\frac{y+4}{-3}=\frac{z+11}{2}$
AnswerCorrect option: B. $\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}$
(b) : DR's of the line joining the given points are
$
\{1-(-3),-2-4,7-11\}
$
i.e., $(4,-6,-4)$ or $(-2,3,2)$
Now, Equation of line passing through $(-3,4,11)$ and having direction ratios $-2,3,2$ is $\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}$
View full question & answer→MCQ 2071 Mark
$P$ is a point on the line segment joining the points $(3,2,-1)$ and $(6,2,-2)$. If $x$ co$-$ordinate of $P$ is $5$ , then its $y$ co$-$ordinate is
AnswerEquation of line joining the points $(3,2,-1)$ and $(6,2,-2)$ is,
$\frac{x-3}{6-3}=\frac{y-2}{2-2}=\frac{z+1}{-2+1}$
i.e., $\frac{x-3}{3}=\frac{y-2}{0}=\frac{z+1}{-1}=\lambda \text { (say) }$
$\Rightarrow x=3 \lambda+3, y=2, z=-\lambda-1$
So, $y-$coordinate of $P$ is $2 .$
View full question & answer→MCQ 2081 Mark
If a line makes angles $\frac{\pi}{2}, \frac{3 \pi}{4}$ and $\frac{\pi}{4}$ with $X, Y$, and $Z$-axes respectively, then its direction cosines are
- ✓
$0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$
- B
$0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$
- C
$0,-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$
- D
$0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$
AnswerCorrect option: A. $0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$
(a) : Here, $l=\cos \frac{\pi}{2}=0$
$
m=\cos \frac{3 \pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)=-\cos \frac{\pi}{4}=\frac{-1}{\sqrt{2}}
$
and $n=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
$\therefore \quad$ Direction cosines are $0, \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.
View full question & answer→MCQ 2091 Mark
Distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is
AnswerCorrect option: D. $\sqrt{\alpha^2+\gamma^2}$
(d) : Foot of perpendicular from $(\alpha, \beta, \gamma)$ on the $y$-axis is $(0, \beta, 0)$.
$\therefore \quad$ Distance of $(\alpha, \beta, \gamma)$ from $y$-axis $=$ distance of $(\alpha, \beta, \gamma)$ from $(0, \beta, 0)$
$
=\sqrt{(0-\alpha)^2+(\beta-\beta)^2+(0-\gamma)^2}=\sqrt{\alpha^2+\gamma^2}
$
View full question & answer→MCQ 2101 Mark
Find the direction cosines of the line $\frac{x-2}{2}=\frac{2 y-5}{-3}=\frac{z+1}{0}$.
AnswerCorrect option: D. $\frac{4}{5},-\frac{3}{5}, 0$
$\frac{x-2}{2}=\frac{2 y-5}{-3}=\frac{z+1}{0}$
$\Rightarrow \frac{x-2}{2}=\frac{y-5 / 2}{-3 / 2}=\frac{z+1}{0}$
Direction cosines are
$\frac{2}{\sqrt{2^2+\left(\frac{-3}{2}\right)^2+0^2}}, \frac{-3 / 2}{\sqrt{2^2+\left(\frac{-3}{2}\right)^2+0^2}}, \frac{0}{\sqrt{2^2+\left(\frac{-3}{2}\right)^2+0^2}}$
i.e., $\frac{2}{5 / 2}, \frac{-3 / 2}{5 / 2}, 0$
i.e., $\frac{4}{5}, \frac{-3}{5}, 0$
View full question & answer→MCQ 2111 Mark
The equation of a line passing through the point $(-3,2,-4)$ and equally inclined to the axes are
AnswerCorrect option: B. $x+3=y-2=z+4$
Since, line equally inclined to the axes.
$\therefore l=m=n ...(i)$
The required equation of line is
$\frac{x+3}{l}=\frac{y-2}{l}=\frac{z+4}{l}\ [$Using $(i)] $
$\Rightarrow \frac{x+3}{1}=\frac{y-2}{1}=\frac{z+4}{1} $
$\Rightarrow x+3=y-2=z+4$
View full question & answer→MCQ 2121 Mark
The cartesian equation of a line is $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$. The vector equation for the line is
- A
$2 \hat{i}+3 \hat{j}-6 \hat{k}+\lambda(2 \hat{i}-3 \hat{j}+2 \hat{k})$
- ✓
$-3 \hat{i}+5 \hat{j}-6 \hat{k}+\lambda(2 \hat{i}+4 \hat{j}+2 \hat{k})$
- C
$-3 \hat{i}-5 \hat{j}+6 \hat{k}+\lambda(2 \hat{i}-3 \hat{j}-2 \hat{k})$
- D
$3 \hat{i}+5 \hat{j}+6 \hat{k}+\lambda(2 \hat{i}-4 \hat{j}-2 \hat{k})$
AnswerCorrect option: B. $-3 \hat{i}+5 \hat{j}-6 \hat{k}+\lambda(2 \hat{i}+4 \hat{j}+2 \hat{k})$
(b) : The given cartesian equation is
$
\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2} \text {. }
$
The line passes through the point $(-3,5,-6)$ and is parallel to vector $2 \hat{i}+4 \hat{j}+2 \hat{k}$.
Hence, the vector equation of the line is
$
\vec{r}=-3 \hat{i}+5 \hat{j}-6 \hat{k}+\lambda(2 \hat{i}+4 \hat{j}+2 \hat{k}) .
$
View full question & answer→MCQ 2131 Mark
Find the equation of a line passing through $(1,2,-3)$ and parallel to the line $\frac{x-2}{1}=\frac{y+1}{3}=\frac{z-1}{4}$.
- A
$\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z-1}{1}$
- ✓
$\frac{x-1}{1}=\frac{y-2}{3}=\frac{z+3}{4}$
- C
$\frac{x+1}{1}=\frac{y-2}{3}=\frac{z+3}{4}$
- D
$\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z-1}{1}$
AnswerCorrect option: B. $\frac{x-1}{1}=\frac{y-2}{3}=\frac{z+3}{4}$
(b) : Since, the line is parallel to the line
$
\frac{x-2}{1}=\frac{y+1}{3}=\frac{z-1}{4} \text {. }
$
$\therefore \quad$ D.r.'s of the required line are < 1,3,4 >.
Hence, equation of the line passing through $(1,2,-3)$ with d.r.'s < 1,3,4 > is $\frac{x-1}{1}=\frac{y-2}{3}=\frac{z+3}{4}$
View full question & answer→MCQ 2141 Mark
A line makes angles $\alpha, \beta$ and $\gamma$ with the co$-$ordinate axes. If $\alpha+\beta=90^{\circ}$, then the value of angle $\gamma$ is
- A
$60^{\circ}$
- ✓
$90^{\circ}$
- C
$45^{\circ}$
- D
$30^{\circ}$
AnswerCorrect option: B. $90^{\circ}$
We know that $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$\Rightarrow \cos ^2 \alpha+\cos ^2\left(90^{\circ}-\alpha\right)+\cos ^2 \gamma=1 \quad\left[\because \alpha+\beta=90^{\circ}\right]$
$\Rightarrow \cos ^2 \alpha+\sin ^2 \alpha+\cos ^2 \gamma=1$
$\Rightarrow 1+\cos ^2 \gamma=1$
$\Rightarrow \cos ^2 \gamma=0$
$\Rightarrow \cos \gamma=0$
$\Rightarrow \gamma=\frac{\pi}{2}$
$=90^{\circ}$
View full question & answer→MCQ 2151 Mark
The equation of a line is given by $\frac{4-x}{2}=\frac{y+3}{3}=\frac{z+2}{6}$, the direction cosines of line parallel to the given line is
- A
$\frac{-2}{7}, \frac{-3}{7}, \frac{-6}{7}$
- B
$\frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}$
- C
$\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$
- ✓
$\frac{-2}{7}, \frac{3}{7}, \frac{6}{7}$
AnswerCorrect option: D. $\frac{-2}{7}, \frac{3}{7}, \frac{6}{7}$
Equation of given line is $\frac{4-x}{2}=\frac{y+3}{3}=\frac{z+2}{6}$.
The direction ratios of the given line are $-2,3,6$.
$\therefore \quad$ The direction cosines of the given line are
$\left(\frac{-2}{\sqrt{4+9+36}}, \frac{3}{\sqrt{4+9+36}}, \frac{6}{\sqrt{4+9+36}}\right)$
$=\left(\frac{-2}{\sqrt{49}}, \frac{3}{\sqrt{49}}, \frac{6}{\sqrt{49}}\right)$
$=\left(\frac{-2}{7}, \frac{3}{7}, \frac{6}{7}\right)$
View full question & answer→MCQ 2161 Mark
If lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}$ $=\frac{z-6}{-5}$ are mutually perpendicular, then $k$ is equal to
- ✓
$-\frac{10}{7}$
- B
$-\frac{7}{10}$
- C
- D
AnswerCorrect option: A. $-\frac{10}{7}$
(a) : Lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are perpendicular if $a_1 a_2+b_1 b_2+c_1 c_2=0$.
$
\Rightarrow-3(3 k)+2 k+2(-5)=0 \Rightarrow k=-\frac{10}{7}
$
View full question & answer→MCQ 2171 Mark
Write the direction cosines of a line parallel to the line $\frac{3-x}{3}=\frac{y+2}{-2}=\frac{z+2}{6}$.
- A
$\frac{1}{7}, \frac{2}{7}, \frac{3}{7}$
- ✓
$\frac{-3}{7}, \frac{-2}{7}, \frac{6}{7}$
- C
$\frac{3}{7}, \frac{2}{7}, \frac{6}{7}$
- D
$\frac{3}{7}, \frac{-2}{7}, \frac{6}{7}$
AnswerCorrect option: B. $\frac{-3}{7}, \frac{-2}{7}, \frac{6}{7}$
(b) : We have, $\frac{x-3}{-3}=\frac{y+2}{-2}=\frac{z+2}{6}$
$\Rightarrow$ Direction ratios are $-3,-2,6$.
$\therefore \quad$ Direction cosines are $\frac{-3}{7}, \frac{-2}{7}, \frac{6}{7}$.
These are direction cosines of a line parallel to given line.
View full question & answer→MCQ 2181 Mark
If the lines $\frac{x+2}{4 \lambda+1}=\frac{y-1}{4}=\frac{z}{-18}$ and $\frac{x}{-3}=\frac{y+1}{5 \mu-3}$ $=\frac{z-1}{6}$ are parallel to each other, then the value of the pair $(\lambda, \mu)$ is
- A
$\left(-2, \frac{1}{3}\right)$
- B
$\left(2,-\frac{1}{3}\right)$
- ✓
$\left(2, \frac{1}{3}\right)$
- D
AnswerCorrect option: C. $\left(2, \frac{1}{3}\right)$
Consider, $L_1: \frac{x+2}{4 \lambda+1}=\frac{y-1}{4}=\frac{z}{-18}$ and $L_2: \frac{x}{-3}=\frac{y+1}{5 \mu-3}=\frac{z-1}{6}$
If two lines are parallel, then their direction ratios are proportional.
$\therefore \frac{4 \lambda+1}{-3}=\frac{4}{5 \mu-3}=\frac{-18}{6}$
$\Rightarrow \frac{4 \lambda+1}{-3}=-3$ and $\frac{4}{5 \mu-3}=-3$
$\Rightarrow 4 \lambda+1=9$ and $4=-15 \mu+9$
$\Rightarrow 4 \lambda=8$ and $15 \mu=5$
$\Rightarrow \lambda=2$ and $\mu=\frac{1}{3}$
So, the value of the pair $(\lambda, \mu)$ is $\left(2, \frac{1}{3}\right)$.
View full question & answer→MCQ 2191 Mark
Find the shortest distance between the given two lines :
$\frac{x+1}{1}=\frac{y+1}{-1}=\frac{z+1}{1}$ and $\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{4}$.
- ✓
$\frac{4}{\sqrt{78}}$ units
- B
$\frac{3}{\sqrt{78}}$ units
- C
$\frac{5}{\sqrt{78}}$ units
- D
$\frac{7}{\sqrt{78}}$ units
AnswerCorrect option: A. $\frac{4}{\sqrt{78}}$ units
(a) : $x_1=-1, y_1=-1, z_1=-1, a_1=1, b_1=-1, c_1=1$ and $x_2=2, y_2=3, z_2=4, a_2=2, b_2=3, c_2=4$
$
\begin{aligned}
\therefore \quad & d=\left|\frac{\left|\begin{array}{ccc}
3 & 4 & 5 \\
1 & -1 & 1 \\
2 & 3 & 4
\end{array}\right|}{\sqrt{(-4-3)^2+(2-4)^2+(3+2)^2}}\right| \\
& =\left|\frac{3(-4-3)-4(4-2)+5(3+2)}{\sqrt{49+4+25}}\right| \\
& =\left|\frac{-21-8+25}{\sqrt{78}}\right|=\frac{4}{\sqrt{78}} \text { units }
\end{aligned}
$
View full question & answer→MCQ 2201 Mark
If $\left(\frac{1}{2}, \frac{1}{3}, n\right)$ are the direction cosines of a line, then the value of $n$ is
AnswerCorrect option: A. $\pm \frac{\sqrt{23}}{6}$
(a) : $\because\left(\frac{1}{2}, \frac{1}{3}, n\right)$ are the direction cosines of a line
$
\therefore\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+n^2=1 \Rightarrow n^2=\frac{23}{36} \Rightarrow n=\frac{ \pm \sqrt{23}}{6}
$
View full question & answer→MCQ 2211 Mark
Find the equation of a line passing through a point $(2,-1,3)$ and parallel to the line $\vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}+\hat{j}-2 \hat{k})$.
- A
$\vec{r}=(\hat{i}+\hat{j})+\mu(2 \hat{i}-\hat{j}+3 \hat{k})$
- ✓
$\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\mu(2 \hat{i}+\hat{j}-2 \hat{k})$
- C
$\vec{r}=(\hat{i}-\hat{j})+\mu(2 \hat{i}-\hat{j}+3 \hat{k})$
- D
$\vec{r}=(2 \hat{i}+\hat{j}+3 \hat{k})+\mu(2 \hat{i}+\hat{j}-2 \hat{k})$
AnswerCorrect option: B. $\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\mu(2 \hat{i}+\hat{j}-2 \hat{k})$
(b) : The given line is parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$ and the required line is parallel to the given line. So, required line is parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$. Thus, the equation of the required line passing through (2, $-1,3)$ is $\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\mu(2 \hat{i}+\hat{j}-2 \hat{k})$
View full question & answer→MCQ 2221 Mark
Find the direction cosines of the line joining $A(0,7,10)$ and $B(-1,6,6)$.
- ✓
$\left(\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{4}{3 \sqrt{2}}\right)$
- B
$\left(\frac{1}{3 \sqrt{2}}, \frac{4}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)$
- C
$\left(\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)$
- D
$\left(\frac{4}{3 \sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)$
AnswerCorrect option: A. $\left(\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{4}{3 \sqrt{2}}\right)$
(a) : Direction ratios of $A B$ are
$(-1-0,6-7,6-10)$ or $(-1,-1,-4)$
Also, $\sqrt{(-1)^2+(-1)^2+(-4)^2}=3 \sqrt{2}$
$\therefore \quad$ Direction cosines are $\left(-\frac{1}{3 \sqrt{2}},-\frac{1}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}}\right)$
or $\left(\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{4}{3 \sqrt{2}}\right)$
View full question & answer→MCQ 2231 Mark
Find the direction cosines of the line that makes equal angles with the three axes in space.
- A
$\pm \frac{1}{\sqrt{2}}$
- B
$\pm 1$
- ✓
$\pm \frac{1}{\sqrt{3}}$
- D
$\sqrt{3}$
AnswerCorrect option: C. $\pm \frac{1}{\sqrt{3}}$
(c) : Since $l=m=n$ and $l^2+m^2+n^2=1$
$\Rightarrow l=m=n= \pm \frac{1}{\sqrt{3}}$
View full question & answer→MCQ 2241 Mark
The direction cosines of the line passing through two points $(2,1,0)$ and $(1,-2,3)$ are
- A
$\left\langle\frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}}, \frac{3}{\sqrt{19}}\right\rangle$
- B
$\left\langle\frac{-3}{\sqrt{19}}, \frac{3}{\sqrt{19}}, \frac{-1}{\sqrt{19}}\right\rangle$
- ✓
$\left\langle\frac{-1}{\sqrt{19}}, \frac{-3}{\sqrt{19}}, \frac{3}{\sqrt{19}}\right\rangle$
- D
$\left\langle\frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}}, \frac{-3}{\sqrt{19}}\right\rangle$
AnswerCorrect option: C. $\left\langle\frac{-1}{\sqrt{19}}, \frac{-3}{\sqrt{19}}, \frac{3}{\sqrt{19}}\right\rangle$
(c) : Here, $P(2,1,0)$ and $Q(1,-2,3)$
So, $P Q=\sqrt{(1-2)^2+(-2-1)^2+(3-0)^2}$ $=\sqrt{1+9+9}=\sqrt{19}$
Thus, the direction cosines of the line joining two points are $\left\langle\frac{1-2}{\sqrt{19}}, \frac{-2-1}{\sqrt{19}}, \frac{3-0}{\sqrt{19}}\right\rangle=\left\langle\frac{-1}{\sqrt{19}}, \frac{-3}{\sqrt{19}}, \frac{3}{\sqrt{19}}\right\rangle$
View full question & answer→MCQ 2251 Mark
If a line makes angles $90^{\circ}, 60^{\circ}$ and $30^{\circ}$ with the positive directions of $x, y$ and $z$-axis respectively, then its direction cosines are
- A
$\left\langle\frac{1}{2}, 0, \frac{\sqrt{3}}{2}\right\rangle$
- B
$\left\langle\frac{\sqrt{3}}{2}, \frac{1}{2}, 0\right\rangle$
- C
$\left\langle\frac{\sqrt{3}}{2}, 0, \frac{1}{2}\right\rangle$
- ✓
$\left\langle 0, \frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle$
AnswerCorrect option: D. $\left\langle 0, \frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle$
(d) : Let the direction cosines of the line be $l, m, n$. Then, $l=\cos 90^{\circ}=0, m=\cos 60^{\circ}=\frac{1}{2}$ and $n=\cos 30^{\circ}=\frac{\sqrt{3}}{2}$. So, direction cosines are $\left\langle 0, \frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle$.
View full question & answer→MCQ 2261 Mark
If $\alpha, \beta, \gamma$ are the direction angles of a vector and $\cos \alpha=\frac{14}{15}, \cos \beta=\frac{1}{3}$, then $\cos \gamma=$
- ✓
$\pm \frac{2}{15}$
- B
$\pm \frac{1}{5}$
- C
$\pm \frac{1}{15}$
- D
$\pm \frac{4}{15}$
AnswerCorrect option: A. $\pm \frac{2}{15}$
$\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$(\because \alpha, \beta, \gamma$ are direction angles)
$\Rightarrow \frac{196}{225}+\frac{1}{9}+\cos ^2 \gamma=1$
$\Rightarrow \cos ^2 \gamma=1-\frac{221}{225}=\frac{4}{225}$
$\Rightarrow \cos \gamma= \pm \frac{2}{15}$
View full question & answer→MCQ 2271 Mark
If $\alpha, \beta, \gamma$ are the angles made by a line with the co$-$ordinate axes. Then $\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma$ is
Answer$\because \alpha, \beta$ and $\gamma$ are the angles made by line with the co$-$ordinate axes.
$\therefore \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$\Rightarrow 1-\sin ^2 \alpha+1-\sin ^2 \beta+1-\sin ^2 \gamma=1$
$\Rightarrow \sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=2$
View full question & answer→MCQ 2281 Mark
If the equation of a line $A B$ is $\frac{x-3}{1}=\frac{y+2}{-2}$ $=\frac{z-5}{4}$, find the direction ratios of a line parallel to $A B$.
- A
$1,2,4$
- B
$1,2,-4$
- C
$1,-2,-4$
- ✓
$1,-2,4$
AnswerCorrect option: D. $1,-2,4$
(d) : The direction ratios of line parallel to $A B$ is $1,-2$ and 4 .
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If a line makes an angle $\theta_1, \theta_2, \theta_3$ with the axes respectively, then the value of $\cos 2 \theta_1+\cos 2 \theta_2$ $+\cos 2 \theta_3$ is
AnswerConsider, $\cos 2 \theta_1+\cos 2 \theta_2+\cos 2 \theta_3$
$=2\left(\cos ^2 \theta_1+\cos ^2 \theta_2+\cos ^2 \theta_3\right)-3 \quad\left(\because \cos 2 x=2 \cos ^2 x-1\right)$
$=2(1)-3=-1$
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