Find the value of $k$ for which the lines $x - 2 = 2y + 1 = -z + 1$ and $x = -y = kz$ are perpendicular to each other.

Find the value of k for which the lines x - 2 = 2y + 1 — CBSE Class 12 Mathematics 2020 PYQ

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CBSECBSE Class XII Board 2020Subjective· 2mImportance★★★★★

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The key idea is that two lines are perpendicular when the dot product of their direction vectors is zero. For the given lines, this condition yields $k = -2$ .

Why Direction Vectors?

Two lines in 3D are perpendicular if and only if their direction vectors are perpendicular — that is, their dot product equals zero. The given equations are in symmetric form, which directly gives us the direction ratios. The trick is to rewrite each line so the direction numbers are clear.

Step-by-step solution

1. Find the direction vector of the first line.

The line is $x - 2 = 2y + 1 = -z + 1$ . To get it into standard symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ , we need each expression to have coefficient $1$ for the variable.

Rewrite $2y + 1$ as $2\left(y + \frac12\right)$ . So the common value becomes:

x - 2 = 2\left(y + \frac12\right) = -z + 1

Now divide the middle term by $2$ to isolate $y + \frac12$ :

x - 2 = \frac{y + \frac12}{\frac12} = -z + 1

For the $z$ term, note $-z + 1 = -(z - 1)$ . So we can write:

x - 2 = \frac{y + \frac12}{\frac12} = \frac{z - 1}{-1}

Thus the direction ratios are the denominators: $(1, \frac12, -1)$ . To avoid fractions, multiply through by $2$ — direction ratios can be scaled arbitrarily. So the direction vector of the first line is:

\vec{d_1} = (2, 1, -2)

Tip

Always clear denominators in direction ratios — it makes dot products cleaner and reduces fraction errors.

2. Find the direction vector of the second line.

The line is $x = -y = kz$ . Write each as a fraction with denominator $1$ :

\frac{x}{1} = \frac{y}{-1} = \frac{z}{1/k}

So the direction ratios are $(1, -1, \frac1k)$ . Multiply through by $k$ to clear the fraction:

$\vec{d_2} = (k, -k, 1)$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 65/2/11 markMCQ
$x$
›Reveal solutionSolution
The vector makes equal angles with the $x$ and $y$ axes and is perpendicular to the $z$ axis, so its $z$ -component is zero and its $x$ and $y$ components are equal. With magnitude 3, each component is $\frac{3}{\sqrt{2}}$ , giving the vector $\frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}$ , which matches option (C) after rationalising.

The key idea here is that a vector’s direction cosines tell you how it’s oriented relative to the axes. If a vector makes equal angles with the $x$ and $y$ axes, its direction cosines (and hence its components) along those axes are equal. And if it’s perpendicular to the $z$ axis, its $z$ -component is zero — the vector lies entirely in the $xy$ -plane.

So we’re looking for a vector of the form $a\hat{i} + a\hat{j} + 0\hat{k}$ , where $a$ is the (equal) $x$ and $y$ components. The magnitude condition then fixes $a$ .

Let’s work through it.

Set up the vector form.

Let the vector be $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ .

Perpendicular to the $z$ axis means $v_z = 0$ .

Equal angles with $x$ and $y$ axes means the direction cosines $\cos\alpha$ and $\cos\beta$ are equal. Since direction cosines are proportional to the components, we have $v_x = v_y$ .

So $\vec{v} = a\hat{i} + a\hat{j}$ , where $a = v_x = v_y$ .

Apply the magnitude condition.

The magnitude is given as 3:

$|\vec{v}| = \sqrt{a^2 + a^2 + 0^2} = \sqrt{2a^2} = |a|\sqrt{2} = 3.$
Since magnitude is positive, $a$ must be positive (the vector’s direction is fixed by the signs, but the problem doesn’t specify a sign, so we take the positive one).
$a\sqrt{2} = 3 \quad\Rightarrow\quad a = \frac{3}{\sqrt{2}}.$

Write the vector.

$\vec{v} = \frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}.$
This is not yet in the form of any option — but option (C) is $\frac{3\sqrt{2}}{2} \hat{i} + \frac{3\sqrt{2}}{2} \hat{j}$ . Notice that $\frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ (multiply numerator and denominator by $\sqrt{2}$ ). So they are identical. …
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CBSE 2026Set V11 markMCQ
$x$
(a) $0, k, 0$
(b) $0, 0, k$
(c) $k, 0, 0$
(d) $k, k, k$
›Reveal solutionSolution

The $x$ -axis points along $\hat i=(1,0,0)$ , so its direction ratios are $k,0,0$ ; answer (c).

The positive $x$ -axis is directed along the unit vector $\hat i=(1,0,0)$ . Direction ratios are any nonzero scalar multiple of the d …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Direction ratios of the straight line vector r = 2i - 3j + k + m(9i - 2j + 5k) are:$
(a) <2, -3, 1>
(b) <9, 2, 5>
(c) <-2, 3, -1>
(d) <9, -2, 5>
›Reveal solutionSolution

In the vector form $\vec r = \vec a + m\vec b$ , the coefficients of the parameter $m$ (i.e. $\vec b$ ) directly give the direction ratios of the line.

The line is $\vec r = 2\hat i - 3\hat j + \hat k + m(9\hat i - 2\hat j + 5\hat k)$ .

…

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CBSE 2025Set ANNUAL1 mark
$Q. Direction ratios of straight line r⃗ = 4î - 7ĵ + 5k̂ + s(9î - 2ĵ + 5k̂) are ______.$
›Reveal solutionSolution

In the vector form $\vec r = \vec a + s\,\vec b$ , the coefficients of the parameter give the line's direction ratios.

For $\vec r = 4\hat i - 7\hat j + 5\hat k + s(9\hat i - 2\hat j + 5\hat k)$ , the fixed point is $(4,-7,5)$ and the direction vector is …

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CBSE 2024Set ANNUAL1 mark
$(2, 3, -4)$
›Reveal solutionSolution

Direction ratios of a line joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are $(x_2-x_1,\,y_2-y_1,\,z_2-z_1)$ .

Let the points be $P(2,3,-4)$ and $Q(1,3,-2)$ .

Direction ratios of line PQ: …

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CBSE 2022Set ANNUAL1 markMCQ
$3x + 4y + 5z - 6 = 0$
(a) $3, 4, 5$
(b) $-3, 4, 5$
(c) $3, -4, 5$
(d) $2, 3, -4$
›Reveal solutionSolution

For $ax+by+cz+d=0$ , the normal has direction ratios $(a,b,c)=(3,4,5)$ .

The normal to a plane $ax+by+cz+d=0$ has direction ratios equal to the coefficients $(a,b,c)$ …

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CBSE 2019Set HE1 mark
$Q. Fill in the blank: Direction ratio of two parallel lines will be ______.$
›Reveal solutionSolution

Parallel lines have the same direction, so their direction ratios are proportional.

Direction ratios $(a,b,c)$ describe the direction a line points in. Two lines are parallel precisely when they point in the same (or exactly opposite) direction, which means one line's direction ratios are a scalar multiple of the oth …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Direction ratios of the line joining points A(2, 3, -4) and B(1, -2, 3) is:$
(a)
(i) -1, 5, 7
(b)
(ii) -1, -5, 7
(c)
(iii) 1, -5, -7
(d)
(iv) -1, -5, 7
›Reveal solutionSolution

Direction ratios $=(-1,-5,7)$ — option (iv).

Concept. For a line joining $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ , the direction ratios are $x_2-x_1,\ y_2-y_1,\ z_2-z_1$ .

Steps. With $A(2,3,-4)$ and $B(1,-2,3)$ : …

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Q.Find the value of $k$ for which the lines $x - 2 = 2y + 1 = -z + 1$ and $x = -y = kz$ are perpendicular to each other.

Concept understanding — Direction Vectors

Direction Vectors

The Idea

Vector Equation of a Line

Direction Ratios and Direction Cosines

Why It Matters …

Why Direction Vectors?

Step-by-step solution

Q.Find the value of kkk for which the lines x−2=2y+1=−z+1x - 2 = 2y + 1 = -z + 1x−2=2y+1=−z+1 and x=−y=kzx = -y = kzx=−y=kz are perpendicular to each other.

Why Direction Vectors?

Step-by-step solution

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Q.Find the value of $k$ for which the lines $x - 2 = 2y + 1 = -z + 1$ and $x = -y = kz$ are perpendicular to each other.

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