CBSE Sample Paper 2022-23 · Section B · 2 marks

Q23: $\lambda$ for $\vec a + \vec b \perp \vec a - \vec b$ ⁠, or direction cosines from $6x - 12 = 3y + 9 = 2z - 2$

Sample 2022-23Ch 10 · Vector Algebra Ch 11 · Three Dimensional Geometry

The question

(a) If $\vec a = \hat i - \hat j + 7\hat k$ and $\vec b = 5\hat i - \hat j + \lambda\hat k$ ⁠, then find the value of $\lambda$ so that the vectors $\vec a + \vec b$ and $\vec a - \vec b$ are orthogonal.

(b) Find the direction ratio and direction cosines of a line parallel to the line whose equations are

6x - 12 = 3y + 9 = 2z - 2

Exam answer

(a)

\begin{aligned} \vec a + \vec b &= 6\hat i - 2\hat j + (7 + \lambda)\hat k \\ \vec a - \vec b &= -4\hat i + (7 - \lambda)\hat k \end{aligned}

Why?

Add or subtract the matching components:

\begin{aligned} \hat i&:\ 1 + 5 = 6,\ 1 - 5 = -4 \\ \hat j&:\ -1 + (-1) = -2 \\ \hat j&:\ -1 - (-1) = 0 \\ \hat k&:\ 7 + \lambda,\ 7 - \lambda \end{aligned}

$\vec a + \vec b$ and $\vec a - \vec b$ are orthogonal if $(\vec a + \vec b)\cdot(\vec a - \vec b) = 0$ ⁠, i.e. if

\begin{aligned} &6(-4) + (-2)(0) \\ &\quad + (7 + \lambda)(7 - \lambda) = 0 \\ \Rightarrow\ &-24 + 49 - \lambda^2 = 0 \\ \Rightarrow\ &\lambda^2 = 25 \end{aligned}

Why?

Orthogonal means perpendicular, and two non-zero vectors are perpendicular exactly when their dot product is 0. The dot product multiplies matching components and adds:

\vec p\cdot\vec q = p_1q_1 + p_2q_2 + p_3q_3

The last product is a difference of squares: $(7 + \lambda)(7 - \lambda) = 49 - \lambda^2$ ⁠.

\begin{aligned} 6(-4) &= -24 \\ (-2)(0) &= 0 \\ (7 + \lambda)(7 - \lambda) &= 7^2 - \lambda^2 \\ &= 49 - \lambda^2 \end{aligned}

Now solve, one move at a time:

\begin{aligned} -24 + 0 + 49 - \lambda^2 &= 0 \\ 25 - \lambda^2 &= 0 \\ 25 &= \lambda^2 \\ \lambda &= \pm\sqrt{25} = \pm 5 \end{aligned}

Both signs work, because $\lambda$ appears only as $\lambda^2$ ⁠.

$\therefore\ \lambda = \pm 5$

(b) The equations of the line are

\begin{aligned} &6x - 12 = 3y + 9 = 2z - 2 \\ \Rightarrow\ &6(x - 2) = 3(y + 3) = 2(z - 1) \end{aligned}

Why?

To read off direction ratios, the line must be in the standard symmetric form

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

where $x$ ⁠, $y$ ⁠, $z$ each have coefficient 1 on top. So first take the coefficient of each variable outside a bracket.

\begin{aligned} 6x - 12 &= 6x - 6(2) = 6(x - 2) \\ 3y + 9 &= 3y + 3(3) = 3(y + 3) \\ 2z - 2 &= 2z - 2(1) = 2(z - 1) \end{aligned}

Writing $6(x - 2) = \frac{x - 2}{1/6}$ ⁠, and so on, the standard symmetric form is

\frac{x - 2}{\frac16} = \frac{y - (-3)}{\frac13} = \frac{z - 1}{\frac12}

Why?

Multiplying by 6 is the same as dividing by $\frac16$ ⁠. In the same way $3(y + 3) = \frac{y + 3}{1/3}$ and $2(z - 1) = \frac{z - 1}{1/2}$ ⁠.

So the line passes through $(2, -3, 1)$ ⁠, and its direction ratios are the three denominators.

Since the lines are parallel, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ ⁠. Hence, the required direction ratios are $\frac16, \frac13, \frac12$ ⁠, or $1, 2, 3$ (multiplying by 6).

Why?

Parallel lines have the same direction, so their direction ratios are proportional. Any non-zero multiple of a set of direction ratios is again a set of direction ratios. Multiplying by 6, the LCM of the denominators, clears the fractions.

\begin{aligned} \frac16 \times 6 &= 1 \\ \frac13 \times 6 &= 2 \\ \frac12 \times 6 &= 3 \end{aligned}

\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}

Why?

Direction cosines come from direction ratios $a$ ⁠, $b$ ⁠, $c$ by dividing each by

\sqrt{a^2 + b^2 + c^2}

Here $a, b, c = 1, 2, 3$ ⁠:

\begin{aligned} &1^2 + 2^2 + 3^2 \\ &= 1 + 4 + 9 = 14 \\[4pt] &l = \frac{1}{\sqrt{14}},\ m = \frac{2}{\sqrt{14}},\ n = \frac{3}{\sqrt{14}} \end{aligned}

Check: the squares of the direction cosines add up to 1, since $\frac{1 + 4 + 9}{14} = 1$ ⁠.

$\therefore$ The required direction cosines are $\dfrac{1}{\sqrt{14}}, \dfrac{2}{\sqrt{14}}, \dfrac{3}{\sqrt{14}}$ ⁠.

Where this comes from

Chapter 10, Vector Algebra: Orthogonal vectors; direction ratios of a line.

Question type: (a) Perpendicular vectors; (b) From Cartesian to vector form + Direction cosines from angles or from ratios (combined)

How to spot it: click to see

(a) Ch 10, type 5. Perpendicular vectors: set the dot product to 0

Spot it: "is perpendicular to", with an unknown such as $\lambda$ to find.

Here: $\vec a + \vec b$ and $\vec a - \vec b$ both carry $\lambda$ in the $\hat k$ part. Set their dot product to 0: $-24 + 49 - \lambda^2 = 0$ ⁠, so $\lambda = \pm 5$ ⁠.

(b) Ch 11, 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, $6x - 12 = 3y + 9 = 2z - 2$ does not have $x$ ⁠, $y$ ⁠, $z$ alone on top. Rewrite it as $\frac{x - 2}{1/6} = \frac{y + 3}{1/3} = \frac{z - 1}{1/2}$ and read the direction ratios $\frac16, \frac13, \frac12$ ⁠, that is 1, 2, 3.

(b) Ch 11, 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 a parallel line has the same ratios 1, 2, 3: divide each by $\sqrt{14}$ to get the direction cosines.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 11, Three Dimensional Geometry.

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Vector Algebra in past papers: question 41 of 48 · see them all

Q23: λ\lambdaλ for a⃗+b⃗⊥a⃗−b⃗\vec a + \vec b \perp \vec a - \vec ba+b⊥a−b⁠, or direction cosines from 6x−12=3y+9=2z−26x - 12 = 3y + 9 = 2z - 26x−12=3y+9=2z−2

Q23: $\lambda$ for $\vec a + \vec b \perp \vec a - \vec b$ ⁠, or direction cosines from $6x - 12 = 3y + 9 = 2z - 2$