Maths M.C.Q (1 Marks)

2. Externally in the ratio 2 : 3

Solution:

Let the XY-plane divide the line segment joining points

P(-1, 3, 4) and Q(2, -5, 6) in the ratio k : 1.

Using the section formula, the coordinates of the point of intersection are given by

$\Big(\frac{\text{k}(2)-1}{\text{k}+1},\frac{\text{k}(-5)+3}{\text{k}+1},\frac{\text{k}(6)+4}{\text{k}+1}\Big) $

On the XY-plane, the Z-coordinate of any point is zero.

$\Rightarrow\frac{\text{k}(6)+4}{\text{k}+1}=0$

$\Rightarrow6\text{k}+4=0$

$\Rightarrow\text{k}=\frac{-2}{3}$

Thus, the XY-plane divides the line segment joining the given points in the ratio 2 : 3 externally.

2. Parallel.

1. $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$

Solution

L makes an angle $\frac{\pi}{4}$ with X and Y axis and $\frac{\pi}{2}$

$\therefore$ d.cs are $\Big(\cos\frac{\pi}{34},\cos\frac{\pi}{4},\cos\frac{\pi}{2}\Big)=\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$

3. 0, 1, 0

Solution:

When P moves then x = 0, z = 0 but y is not given. Let y = y Then the coordinates of the point will be (0, y, 0) Now, direction cosines with respect to (0, y, 0) is given by.

$\cos\alpha=\frac{0}{0^2+\text{y}^2+0^2}=\frac{0}{\text{y}}=0$

$\cos\beta=\frac{\text{y}}{0^2+\text{y}^2+0^2}=\frac{\text{y}}{\text{y}}=1$

$\cos\gamma=\frac{{0}}{0^2+\text{y}^2+0^2}=\frac{{0}}{\text{y}}=0$

The direction cosines are 0, 1, 0

4. 0, 1, 0

Solution:

Given, x = z = 0

It represents Z-axis

$\therefore$ Direction cosines = (0, 1, 0)

1. $\frac{2}{7},\frac{-3}{7},\frac{-6}{7}$

1. $\frac{2}{3},\frac{2}{3},-\frac{1}{3}$

Solution:

Consider the problem 

Let l, m, n are direction cosines of the given line.

then as it made an acute angle with x−axis,

Therefore, l > 0

The line passes through (6, −7, −1) and (2, −3, 1)

Therefore, its direction ratios are 

6 − 2, −7 + 3, −1−1 or 2, −2, −1

Hence direction cosines of the line are given by $\frac{2}{3},\frac{2}{3},-\frac{1}{3}.$

4. $\frac{-2}{7},\frac{3}{7},\frac{-6}{7}$

Solution:

Given the points are P(1, −2, 4) and Q(−1, 1, −2).

Now the direction ratios of the ray PQ are (−1−1, 1 + 2, −2−4) = (−2, 3, −6).

The direction cosines of the line PQ will be

$\bigg(\frac{2}{\sqrt{2^2+3^2+6^2}},\frac{3}{\sqrt{2^2+3^2+6^2}},\frac{-6}{\sqrt{2^2+3^2+6^2}}\bigg)=\Big(\frac{-2}{7},\frac{3}{7},\frac{-6}{7}\Big).$

3. $\frac{2\sqrt{3}}{5}$

3. $\sqrt{3}$

Solution:

D.C of the line are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Any point on the line at a distance tt from P(2, -1, 2) is

$\Big(2+\frac{\text{t}}{\sqrt{3}},-1+\frac{\text{t}}{\sqrt{3}},2+\frac{\text{t}}{\sqrt{3}}\Big)$

which lies on $2\text{x} + \text{y + z} = 9$

$\Rightarrow\text{t}=\sqrt{3}$

4. $\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$

Solution:

P(1, -2, 4), Q(-1, 1, -2)

$\text{PQ}=\sqrt{(1-(1))^2 +(2-1)^2+(4-(-2))^{2}}$

$=\sqrt{4+9+36}$

$=\sqrt{49}=7\text{DC}$

$=\Big(\frac{-1-1}{7},\frac{1-(2)}{7},\frac{-2-4}{7}\Big)$

$=\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$

1. $\cos^{-1}\Big(\frac{1}{{3}}\Big)$

Solution:

Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, OP, AR, Consider the diagonals OP and AR.

Direction ratios of OP and AR are proportional to a - 0, a - 0, a - 0 and 0 - a, a - 0, a - 0, e.i. a, a, a and -a, a, a, respectivelly.

Let $\theta$ be the angle between OP and AR. Then,

$\cos\theta=\frac{\text{a}\times-\text{a}+\text{a}\times\text{a}+\text{a}\times\text{a}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{(-\text{a})^2+\text{a}^2+\text{a}^2}}$

$\Rightarrow\cos\theta=\frac{-\text{a}+\text{a}^2+\text{a}^2}{\sqrt{3\text{a}^2}\sqrt{3\text{a}^2}}$

$\Rightarrow\cos\theta=\frac{1}{3}$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{3}\Big)$

Similarly, the angles between other pairs of the diagonals are equal to $\cos^{-1}\Big(\frac{1}{3}\Big)$ as the angle between any two diagonals.

2. a, 1, c

Solution:

Given x = ay + b and z = cy + d

$\Rightarrow\frac{\text{x}-\text{b}}{\text{a}}=\text{y}$​ and $\frac{\text{z}-\text{d}}{\text{c}}=\text{y}$

$\Rightarrow\frac{\text{x}-\text{b}}{\text{a}}=\frac{\text{y}}{1}=\frac{\text{z}-\text{d}}{c}$

Therefore Drs of given line is a, 1, c

4. $\frac{\sqrt{2}}{10}$

Solution:

We have, the equation of line as

$\frac{\text{x}-2}{3}=\frac{\text{y}-2}{3}=\frac{\text{z}-2}{3}$

This line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$

Equation of plane is $2\text{x}-2\text{y}+\text{z}=5$

Normal to the plane $\vec{\text{n}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$

Its angle between line and plane is $'\theta'.$

Then $\sin\theta=\frac{|\vec{\text{b}}\cdot{\vec{\text{b}}}|}{|{\vec{\text{b}}}||{\vec{\text{b}}}|}$

$=\frac{\big|(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}})\cdot(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})\big|}{\sqrt{3^2+4^2+5^2}\sqrt{4+4+1}}$

$=\frac{|6-8+5|}{\sqrt{50}\sqrt{9}}$

$=\frac{3}{15\sqrt{2}}=\frac{1}{5\sqrt{2}}$

$\sin\theta=\frac{\sqrt{2}}{10}$

1. $\frac{\pi}{3}$