Board 2023 Class 12 Maths Q11 Solution: Direction cosines of (x − 1)/2 = (1 − y)/3 = (2z − 1)/12

The question

Direction cosines of the line $\frac{x - 1}{2} = \frac{1 - y}{3} = \frac{2z - 1}{12}$ are :

(a) $\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$

(b) $\frac{2}{\sqrt{157}}, -\frac{3}{\sqrt{157}}, \frac{12}{\sqrt{157}}$

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

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

Exam answer

Writing each fraction in standard form:

\begin{aligned} \frac{1 - y}{3} &= \frac{y - 1}{-3} \\ \frac{2z - 1}{12} &= \frac{2\left(z - \frac{1}{2}\right)}{12} = \frac{z - \frac{1}{2}}{6} \end{aligned}

Why?

The standard form of a line is

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

with $x$ ⁠, $y$ ⁠, $z$ each having coefficient $+1$ on top. Only then are $a, b, c$ direction ratios.

So turn $1 - y$ into $-(y - 1)$ ⁠, moving the minus into the denominator, and take the 2 out of $2z - 1$ ⁠, cancelling it against 12.

\begin{aligned} \frac{1 - y}{3} &= \frac{-(y - 1)}{3} \\ &= \frac{y - 1}{-3} \end{aligned}

For $z$ ⁠: $2 \times \frac12 = 1$ ⁠, so $2z - 1 = 2\left(z - \frac12\right)$ ⁠. Then divide top and bottom by 2:

\begin{aligned} \frac{2\left(z - \frac12\right)}{12} &= \frac{z - \frac12}{12 \div 2} \\ &= \frac{z - \frac12}{6} \end{aligned}

In standard form, the line is

\frac{x - 1}{2} = \frac{y - 1}{-3} = \frac{z - \frac{1}{2}}{6}

Direction ratios: $2, -3, 6$ ⁠.

\begin{aligned} &\sqrt{2^2 + (-3)^2 + 6^2} \\ &= \sqrt{4 + 9 + 36} = 7 \end{aligned}

Why?

The direction ratios are the denominators of the standard form. Square each one, add, then take the square root:

\begin{aligned} 2^2 &= 4 \\ (-3)^2 &= (-3)(-3) = 9 \\ 6^2 &= 36 \\[4pt] 4 + 9 &= 13 \\ 13 + 36 &= 49 \\ \sqrt{49} &= 7 \end{aligned}

Dividing each direction ratio by 7, the direction cosines are

\frac{2}{7},\ -\frac{3}{7},\ \frac{6}{7}

Why?

If the direction ratios are $a, b, c$ ⁠, the direction cosines are each divided by $\sqrt{a^2 + b^2 + c^2}$ ⁠:

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

and the same for $m$ and $n$ ⁠. Then $l^2 + m^2 + n^2 = 1$ ⁠.

\begin{aligned} l &= \frac{2}{7},\ m = \frac{-3}{7},\ n = \frac{6}{7} \\[4pt] &l^2 + m^2 + n^2 \\ &= \frac{4}{49} + \frac{9}{49} + \frac{36}{49} \\ &= \frac{49}{49} = 1\ \checkmark \end{aligned}

Watch out: option (a) loses the minus sign from $1 - y$ ⁠, and option (b) uses 12 without first taking out the 2 in $2z - 1$ ⁠.

Answer: (d) $\dfrac{2}{7}, -\dfrac{3}{7}, \dfrac{6}{7}$

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, $\frac{1 - y}{3}$ and $\frac{2z - 1}{12}$ do not start with $y$ or $z$ alone. Rewrite them as $\frac{y - 1}{-3}$ and $\frac{z - \frac12}{6}$ ⁠, so the direction ratios are $2, -3, 6$ ⁠, not $2, 3, 12$ ⁠.

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 $\sqrt{4 + 9 + 36} = 7$ ⁠: divide each ratio by 7 to get $\frac27, -\frac37, \frac67$ ⁠.

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Q11: Direction cosines of x−12=1−y3=2z−112\frac{x - 1}{2} = \frac{1 - y}{3} = \frac{2z - 1}{12}2x−1​=31−y​=122z−1​

Q11: Direction cosines of $\frac{x - 1}{2} = \frac{1 - y}{3} = \frac{2z - 1}{12}$