Sample 2021-22 T2 Class 12 Maths Q4 Solution: Direction cosines of the line (3 − x)/(−1) = (2y − 1)/2 = z/4

The question

Find the direction cosines of the following line:

\frac{3 - x}{-1} = \frac{2y - 1}{2} = \frac{z}{4}

Exam answer

The given line can be written as

\frac{x - 3}{1} = \frac{y - \frac12}{1} = \frac{z}{4}

Why?

The standard form of a line is

\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

with each of $x$ ⁠, $y$ ⁠, $z$ having coefficient 1 on top.

First ratio: multiply its top and bottom by $-1$ ⁠, so $\frac{3 - x}{-1} = \frac{x - 3}{1}$ ⁠. Second ratio: divide its top and bottom by 2, so $\frac{2y - 1}{2} = \frac{y - \frac12}{1}$ ⁠.

\begin{aligned} \frac{3 - x}{-1} &= \frac{(-1)(3 - x)}{(-1)(-1)} \\ &= \frac{-3 + x}{1} = \frac{x - 3}{1} \\[4pt] \frac{2y - 1}{2} &= \frac{2\left(y - \frac12\right)}{2} \\ &= \frac{y - \frac12}{1} \end{aligned}

Watch out: in $\frac{2y - 1}{2}$ the $y$ still has coefficient 2, so that denominator 2 is not a direction ratio.

Its direction ratios are $1, 1, 4$ ⁠.

Why?

In the standard form, the denominators are direction ratios. Any non-zero multiple of them, such as $2, 2, 8$ ⁠, gives the same direction.

\sqrt{1^2 + 1^2 + 4^2} = \sqrt{18} = 3\sqrt2

Why?

Direction cosines are the direction ratios divided by the length $\sqrt{a^2 + b^2 + c^2}$ ⁠:

l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}

and $m$ ⁠, $n$ in the same way with $b$ ⁠, $c$ ⁠. Here $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt2$ ⁠.

\begin{aligned} &\sqrt{1^2 + 1^2 + 4^2} \\ &= \sqrt{1 + 1 + 16} = \sqrt{18} \\ &= \sqrt{9 \times 2} = \sqrt9 \times \sqrt2 = 3\sqrt2 \end{aligned}

Divide each direction ratio by $3\sqrt2$ ⁠:

\begin{aligned} l &= \frac{1}{3\sqrt2} \\ m &= \frac{1}{3\sqrt2} \\ n &= \frac{4}{3\sqrt2} \end{aligned}

Check: $l^2 + m^2 + n^2 = \frac{1 + 1 + 16}{18} = 1$ ⁠, since $(3\sqrt2)^2 = 9 \times 2 = 18$ ⁠. The negatives of all three are also direction cosines of the same line.

$\therefore$ Its direction cosines are $\dfrac{1}{3\sqrt2}, \dfrac{1}{3\sqrt2}, \dfrac{4}{3\sqrt2}$ ⁠.

Where this comes from

Chapter 11, Three Dimensional Geometry: Direction cosines of a line.

Question type: From Cartesian to vector form + Direction cosines from angles or from ratios (combines 2 types)

How to spot it: click to see

Type 5. From Cartesian to vector form: read off the point and the direction

Spot it: a line is given as three equal fractions, and you need its vector form, its point or its direction.

Here: First, $\frac{3 - x}{-1}$ and $\frac{2y - 1}{2}$ do not start with $x$ or $y$ alone. Rewrite them as $\frac{x - 3}{1}$ and $\frac{y - \frac12}{1}$ ⁠, so the direction ratios are $1, 1, 4$ ⁠.

Type 1. Direction cosines from angles or from ratios

Spot it: you are given the angles with the axes, or three direction ratios, and asked for the direction cosines.

Here: Then $\sqrt{1 + 1 + 16} = 3\sqrt2$ ⁠: divide each ratio by it to get $\frac{1}{3\sqrt2}, \frac{1}{3\sqrt2}, \frac{4}{3\sqrt2}$ ⁠.

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Q4: Direction cosines of the line 3−x−1=2y−12=z4\frac{3 - x}{-1} = \frac{2y - 1}{2} = \frac{z}{4}−13−x​=22y−1​=4z​

Q4: Direction cosines of the line $\frac{3 - x}{-1} = \frac{2y - 1}{2} = \frac{z}{4}$