2 Marks Questions

Question 12 Marks

Show that the line through the point $(1,-1,2)$, $(3,4,-2)$ is perpendicular to the line through the point $(0,3,2)$ and $(3,5,6)$.

Answer

Le the given points be $A (1,-1,2)$ and $B (3,4,-2)$. So the direction ratios of line passing through the points A $(1,-1,2)$ and $B (3,4,-2)$ are $3-1,4+1,-2-2$ or $2,5,-4$.
Similarly the other given points are $C(0,3,2)$ and $D(3$, $5,6)$. So the direction ratios of line passing through the points $C (0,3,2)$ and $D (3,5,6)$ are $3-0,5-3,6-2$ or $3,2,4$.
Now of $A B \perp C D$, then
$
a_1 a_2+b_1 b_2+c_1 c_2=0
$
So, $2 \times 3+5 \times 2+(-4) \times 4=6+10-16=0$
Hence the lines passing through given points are perpendicular to each other.

View full question & answer→

Question 22 Marks

Find the angle between the pair of lines given by
$\vec{r}=(3 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})$ and
$\vec{r}=(5 \hat{i}-2 \hat{j})+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})$.

Answer

Let $\theta$ be the angle between vectors $\overrightarrow{b_1}=\hat{i}+2 \hat{j}+2 \hat{k}$
and $\overrightarrow{b_2}=3 \hat{i}+2 \hat{j}+6 \hat{k}$, therefore
$
\begin{aligned}
\cos \theta & =\left|\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}\right|=\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})}{\sqrt{1+4+4} \sqrt{9+4+36}}\right| \\
& =\left|\frac{3+4+12}{3 \times 7}\right|=\frac{19}{21}
\end{aligned}
$
So, $\quad \theta=\cos ^{-1}\left(\frac{19}{21}\right)$

View full question & answer→

Question 32 Marks

Projections of a line on $x, y, z$ axes are respectively 12,5 and $2 \sqrt{14}$. Find the length of the line segment and its direction cosines.

View full question & answer→

Question 42 Marks

If a line makes angle $\alpha, \beta, \gamma$ with the coordinate axes, then find the value of $\cos 2 \alpha+\cos 2 \beta+$ $\cos 2 \gamma$.

View full question & answer→

Question 52 Marks

Find the direction cosines of a line parallel to the given line: $\frac{4-x}{2}=\frac{y+3}{3}=\frac{z+2}{6}$

Answer

The given line is $\frac{4-x}{2} =\frac{y+3}{3}=\frac{z+2}{6}$
$\text { or } \frac{x-4}{-2} =\frac{y+3}{3}=\frac{z+2}{6}$
Hence the direction cosines of the parallel line will be
$\frac{-2}{\sqrt{(-2)^2+(3)^2+(6)^2}}, \frac{3}{\sqrt{(-2)^2+(3)^2+(6)^2}}, \frac{6}{\sqrt{(-2)^2+(3)^2+(6)^2}}$
i.e., $\frac{-2}{7}, \frac{3}{7}, \frac{6}{7}$

View full question & answer→

Maths 2 Marks Questions

Take a timed test