(a) If the lines (x-3)/(1) = (1-y)/(1) = (z+2)/(p) and (2-x)/(3) = (y+1)/(5) = (z+56)/(2p) are mutually perpen

Question

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

(OR)

(b) Find the vector equation of the line passing through the origin and perpendicular to both the lines

\vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k})

and

\vec{r} = \mu(\hat{i} - \hat{j} + \hat{k})

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Part (a)Concept understanding — Mutual Perpendicularity

Perpendicular Vectors: The Dot-Product Test

Two vectors are perpendicular (orthogonal) when they meet at a right angle — like the east and north directions. But you cannot reach for a protractor in 3D, so you need an algebraic test.

The key idea: when two vectors are perpendicular, neither has any "shadow" along the other. Walk along one and you make zero progress in the direction of the other. The dot product measures exactly this overlap, so perpendicularity means the dot product vanishes.

\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0

Why? The dot product has two equal forms:

\vec{a}\cdot\vec{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\vec{a}|\,|\vec{b}|\cos\theta.

When $\theta = 90^\circ$ , $\cos 90^\circ = 0$ , so the product is zero regardless of the vectors' lengths.

Example. For $\vec{a}=(1,2,3)$ and $\vec{b}=(2,-1,0)$ :

\vec{a}\cdot\vec{b} = 1(2)+2(-1)+3(0) = 0,

so they are perpendicular. By contrast $(2,1)\cdot(1,3) = 2+3 = 5 \neq 0$ , so those two are not.

Tip

In 2D, $(x,y)$ and $(y,-x)$ are always perpendicular — swap and negate. To build a vector perpendicular to a given $\vec{a}$ , solve $\vec{a}\cdot\vec{x}=0$ ; there are infinitely many solutions, all lying in the plane across $\vec{a}$ . …

Part (b)Concept understanding — Line Perpendicular To Two Lines

Line Perpendicular to Two Lines

In 3D geometry a very common task is this: two lines are given, and you must find the direction of a third line that is perpendicular to both of them. This shows up when finding a common perpendicular, the shortest distance between two lines, or the normal to a plane containing two directions.

The Core Idea

A line in space is fixed by two things: a point it passes through, and its direction vector. So "find a line perpendicular to two lines" really means "find a vector perpendicular to both of the two given direction vectors."

If the two given lines have direction vectors

\vec{b_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}, \qquad \vec{b_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k},

then a vector perpendicular to both is their cross product:

\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}

Why the Cross Product?

The defining property of $\vec{b_1}\times\vec{b_2}$ is that it is perpendicular to each factor:

(\vec{b_1}\times\vec{b_2})\cdot\vec{b_1} = 0, \qquad (\vec{b_1}\times\vec{b_2})\cdot\vec{b_2} = 0.

So it points in exactly the direction we need — along both perpendicularity conditions at once. This is why one cross product replaces solving a pair of dot-product equations by hand.

Note

The cross product only gives the direction of the perpendicular line. To pin down the actual line you still need a point it must pass through, given by the problem.

Using It

Suppose a line must be perpendicular to $\vec{b_1} = \hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b_2}=\hat{i}-\hat{j}+\hat{k}$ .

\vec{b_1}\times\vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & -1 & 1 \end{vmatrix} = 5\hat{i} + 2\hat{j} - 3\hat{k}.

So the required line has direction ratios $\langle 5, 2, -3\rangle$ . Through a point $A(x_0,y_0,z_0)$ its equation is …

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Two lines are perpendicular when their direction vectors are orthogonal; solving $\vec{b_1}\cdot\vec{b_2}=0$ gives $p=\pm2$ .
A line perpendicular to both given lines has direction $\vec{b_1}\times\vec{b_2}=6\hat i-\hat j-7\hat k$ , so through the origin $\vec r=t(6\hat i-\hat j-7\hat k)$ .

Part (a)

Two lines in space are perpendicular exactly when their direction vectors have zero dot product. First we extract direction ratios by putting each line into the standard form $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$ .

For $\dfrac{x-3}{1}=\dfrac{1-y}{1}=\dfrac{z+2}{p}$ , note $\dfrac{1-y}{1}=\dfrac{y-1}{-1}$ , giving $\vec{b_1}=\hat i-\hat j+p\hat k=(1,-1,p)$ .

For $\dfrac{2-x}{3}=\dfrac{y+1}{5}=\dfrac{z+56}{2p}$ , note $\dfrac{2-x}{3}=\dfrac{x-2}{-3}$ , giving $\vec{b_2}=-3\hat i+5\hat j+2p\hat k=(-3,5,2p)$ .

Watch out

Always factor out the negative sign when a numerator looks like $1-y$ or $2-x$ : $\dfrac{1-y}{1}=\dfrac{y-1}{-1}$ .

Applying the perpendicularity condition:

$\vec{b_1}\cdot\vec{b_2}=(1)(-3)+(-1)(5)+(p)(2p)=-3-5+2p^2=2p^2-8=0.$ …

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

CBSE 2024Set D1 markMCQ
$l, m, n$
(a) $\frac{l}{l_1} = \frac{m}{m_1} = \frac{n}{n_1}$
(b) $\frac{l}{l_1} + \frac{m}{m_1} + \frac{n}{n_1} = 0$
(c) $l^2 + m^2 + n^2 = l_1^2 + m_1^2 + n_1^2$
(d) $ll_1 + mm_1 + nn_1 = 0$
›Reveal solutionSolution

Perpendicular lines $\Rightarrow ll_1 + mm_1 + nn_1 = 0$ .

The angle $\theta$ between two lines with direction ratios $l, m, n$ and $l_1, m_1, n_1$ satisfies

$\cos\theta = \frac{ll_1 + mm_1 + nn_1}{\sqrt{l^2+m^2+n^2}\,\sqrt{l_1^2+m_1^2+n_1^2}}.$ …

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CBSE 2024Set D1 markMCQ
$2x - 4y + 3z = 5$
(a) $-2$
(b) $2$
(c) $3$
(d) none of these
›Reveal solutionSolution

$\lambda = 2$ .

Two planes are perpendicular iff the dot product of their normals is zero. Normals are $\vec{n_1} = (2, -4, 3)$ and $\vec{n_2} = (1, 2, \lambda)$ .

$\vec{n_1}\cdot\vec{n_2} = (2)(1) + (-4)(2) + (3)(\lambda) = 2 - 8 + 3\lambda.$ …

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CBSE 2020Set 65/1/11 markMCQ
$(-1, 5, 4)$
›Reveal solutionSolution
A line perpendicular to the plane $z=0$ must be parallel to the $z$ -axis, so its direction vector is $\hat{k}$ . The line passes through $(-1,5,4)$ , giving $\vec{r} = -\hat{i} + 5\hat{j} + 4\hat{k} + \lambda\hat{k}$ . This matches option (B).

The plane $z = 0$ is the $xy$ -plane — a flat horizontal surface. Any line perpendicular to it must point straight up or down, i.e., parallel to the $z$ -axis. That’s the core geometric insight.

A line’s vector equation is $\vec{r} = \vec{a} + \lambda \vec{d}$ , where $\vec{a}$ is a point on the line and $\vec{d}$ is the direction vector. Here, the direction vector must be along $\hat{k}$ (or any scalar multiple of it). The given point is $(-1, 5, 4)$ , so $\vec{a} = -\hat{i} + 5\hat{j} + 4\hat{k}$ .

Now check each option:

Option (A): $\vec{r} = -\hat{i} + 5\hat{j} + 4\hat{k} + \lambda(\hat{i} + \hat{j})$

Direction is $\hat{i} + \hat{j}$ , which lies in the $xy$ -plane — parallel to $z=0$ , not perpendicular. So this is wrong.

Option (B): $\vec{r} = -\hat{i} + 5\hat{j} + (4 + \lambda)\hat{k}$

Rewrite as $-\hat{i} + 5\hat{j} + 4\hat{k} + \lambda\hat{k}$ . Direction is $\hat{k}$ — exactly what we need. This is correct.

Option (C): $\vec{r} = \hat{i} - 5\hat{j} - 4\hat{k} + \lambda\hat{k}$ …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. The direction cosines of two perpendicular lines are l₁, m₁, n₁ and l₂, m₂, n₂ respectively. Then the direction cosines of a line which is perpendicular to both these lines are:$
(a) l₁ + l₂, m₁ + m₂, n₁ + n₂
(b) l₁ − l₂, m₁ − m₂, n₁ − n₂
(c) m₁n₂ − n₁m₂, n₁l₂ − l₁n₂, l₁m₂ − l₂m₁
(d) m₁n₁ − m₂n₂, l₁n₁ − l₂n₂, l₁m₂ − l₂m₁
›Reveal solutionSolution

A line perpendicular to two given lines is along their cross product.

If two lines have direction ratios (l₁,m₁,n₁) and (l₂,m₂,n₂), a vector perpendicular to both is given by the cross product of the two direction vectors:

(l₁î+m₁ĵ+n₁k̂) × (l₂î+m₂ĵ+n₂k̂) = (m₁n₂−n₁m₂) î + (n₁l₂−l₁n₂) ĵ + (l₁m₂−l₂m₁) k̂

…

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CBSE 2018Set ANNUAL1 markMCQ
$2x - 4y + 3z = 5$
(a) $-2$
(b) $2$
(c) $3$
(d) none of these
›Reveal solutionSolution

Normals must be perpendicular: $2-8+3\lambda=0\Rightarrow\lambda=2$ .

Normals are $\vec n_1=(2,-4,3)$ and $\vec n_2=(1,2,\lambda)$ .

…

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Q.(a) If the lines x−31=1−y1=z+2p\frac{x-3}{1} = \frac{1-y}{1} = \frac{z+2}{p}1x−3​=11−y​=pz+2​ and 2−x3=y+15=z+562p\frac{2-x}{3} = \frac{y+1}{5} = \frac{z+56}{2p}32−x​=5y+1​=2pz+56​ are mutually perpendicular, then find the value(s) of ppp.

Part (a)

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Q.(a) If the lines $\frac{x-3}{1} = \frac{1-y}{1} = \frac{z+2}{p}$ and $\frac{2-x}{3} = \frac{y+1}{5} = \frac{z+56}{2p}$ are mutually perpendicular, then find the value(s) of $p$ .

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