CBSE Class 12 Mathematics Chapter 11: Three Dimensional Geometry NCERT Solutions

Let the direction cosines of the line be $l, m,$ and $n$. These are defined as the cosines of the angles the line makes with the positive x, y, and z axes, respectively.

Given angles are:

• Angle with x-axis, $\alpha = 90^{\circ}$
• Angle with y-axis, $\beta = 135^{\circ}$
• Angle with z-axis, $\gamma = 45^{\circ}$

The direction cosines are calculated as follows:

$l = \cos \alpha = \cos 90^{\circ} = 0$

$m = \cos \beta = \cos 135^{\circ} = \cos (180^{\circ} - 45^{\circ}) = -\cos 45^{\circ} = -\frac{1}{\sqrt{2}}$

$n = \cos \gamma = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$

Thus, the direction cosines of the line are $0, -\frac{1}{\sqrt{2}},$ and $\frac{1}{\sqrt{2}}$.

Let the direction cosines of the line be $l, m,$ and $n$. If the line makes equal angles with each of the coordinate axes, let this angle be $\alpha$. Then, $l = \cos \alpha$, $m = \cos \alpha$, and $n = \cos \alpha$.

We know the fundamental property of direction cosines: $l^2 + m^2 + n^2 = 1$.

Substituting the values, we get:

$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$

$3 \cos^2 \alpha = 1$

$\cos^2 \alpha = \frac{1}{3}$

Taking the square root, we find $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.

Therefore, the direction cosines are $l = \pm \frac{1}{\sqrt{3}}$, $m = \pm \frac{1}{\sqrt{3}}$, and $n = \pm \frac{1}{\sqrt{3}}$. This gives two possible sets of direction cosines: $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ and $(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}})$.

Let the direction ratios of the line be $a = -18$, $b = 12$, and $c = -4$. The direction cosines $(l, m, n)$ can be found using the formula: $l = \frac{a}{\pm\sqrt{a^2+b^2+c^2}}$, $m = \frac{b}{\pm\sqrt{a^2+b^2+c^2}}$, $n = \frac{c}{\pm\sqrt{a^2+b^2+c^2}}$.

First, calculate the magnitude of the direction ratios:

$\sqrt{a^2+b^2+c^2} = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$

$= \sqrt{324 + 144 + 16}$

$= \sqrt{484}$

$= 22$

Now, calculate the direction cosines:

$l = \frac{-18}{\pm 22} = \frac{-18}{22} = -\frac{9}{11}$

$m = \frac{12}{\pm 22} = \frac{12}{22} = \frac{6}{11}$

$n = \frac{-4}{\pm 22} = \frac{-4}{22} = -\frac{2}{11}$

Hence, the direction cosines are $-\frac{9}{11}, \frac{6}{11},$ and $-\frac{2}{11}$.

Let the given points be $A(2,3,4)$, $B(-1,-2,1)$, and $C(5,8,7)$.

To show that these points are collinear, we can check if the direction ratios of the line segment AB are proportional to the direction ratios of the line segment BC.

The direction ratios of a line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

Direction ratios of AB:

$= (-1 - 2, -2 - 3, 1 - 4)$

$= (-3, -5, -3)$

Direction ratios of BC:

$= (5 - (-1), 8 - (-2), 7 - 1)$

$= (5 + 1, 8 + 2, 6)$

$= (6, 10, 6)$

Now, we check if the direction ratios of AB are proportional to the direction ratios of BC. We can see if one set of ratios is a scalar multiple of the other.

Let's compare the ratios: $\frac{6}{-3} = -2$, $\frac{10}{-5} = -2$, $\frac{6}{-3} = -2$.

Since the ratio is constant ($-2$) for all corresponding components, the direction ratios of AB are proportional to the direction ratios of BC. This means that the line segments AB and BC are parallel and share a common point B. Therefore, the points A, B, and C lie on the same straight line, proving they are collinear.