The two lines 𝑥−1 3 = −𝑦 , 𝑧 + 1 = 0 and −𝑥 2 = 𝑦+1 2 = 𝑧 + 2 intersect at a point whose y-coordinate is

Question

Q. The two lines 𝑥-1 3 = -𝑦 , 𝑧 + 1 = 0 and -𝑥 2 = 𝑦+1 2 = 𝑧 + 2 intersect at a point whose y-coordinate is 1. Find the co-ordinates of their point of intersection. Find the vector equation of a line perpendicular to both the given lines and passing through this point of intersection. 5

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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 $\vec{a}$ .

Where it shows up: proving two lines or planes meet at right angles, showing the work done by a force perpendicular to displacement is zero ( $W=\vec{F}\cdot\vec{d}=0$ ), and classic results like "the diagonals of a rhombus are perpendicular." Whenever you read "perpendicular" or "orthogonal," reach for dot product $= 0$ .

This dot-product test for perpendicular vectors is the same foundational NCERT Class 12 Vector Algebra result behind countless CBSE board and JEE Main questions on right angles in 3D. Searches like "how to check if two vectors are perpendicular" consistently lead back to this single condition, which also explains why a force perpendicular to displacement does zero work in Physics.

Concept: Mutual Perpendicularity – a line perpendicular to two given lines is parallel to the cross product of their direction vectors.

Step 1: Write both lines in symmetric form.

First line: $\frac{x-1}{3} = \frac{y}{-1} = \frac{z+1}{0}$ (since $z+1=0$ means $z=-1$ constant).

Direction vector: $\vec{d_1} = (3, -1, 0)$ .

Second line: $\frac{-x}{2} = \frac{y+1}{2} = z+2$ → rewrite as $\frac{x}{-2} = \frac{y+1}{2} = \frac{z+2}{1}$ .

Direction vector: $\vec{d_2} = (-2, 2, 1)$ .

Step 2: Find intersection point with $y=1$ .

From first line: $y = -t$ , so $1 = -t \Rightarrow t = -1$ . Then $x = 1 + 3t = 1 - 3 = -2$ , $z = -1$ .

Check second line: for $t=-1$ , $\frac{x}{-2} = -1 \Rightarrow x=2$ (mismatch). So use second line: let parameter $s$ .

From second line: $y+1 = 2s$ , so $1+1=2s \Rightarrow s=1$ . Then $x = -2s = -2$ , $z+2 = s \Rightarrow z = -1$ .

Consistent: intersection point is $(-2, 1, -1)$ .

Step 3: Direction of perpendicular line.

$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 0 \\ -2 & 2 & 1 \end{vmatrix} = (-1)(1) - (0)(2)\,\mathbf{i} - [3(1)-0(-2)]\,\mathbf{j} + [3(2)-(-1)(-2)]\,\mathbf{k}$

$= (-1)\mathbf{i} - (3)\mathbf{j} + (6-2)\mathbf{k} = (-1, -3, 4)$ .

Vector equation: $\vec{r} = (-2, 1, -1) + \lambda(-1, -3, 4)$ .

✓Final answer

(-2, 1, -1)

The lines meet at $(-2,1,-1)$ ; a line perpendicular to both has direction $\vec d_1\times\vec d_2=-\hat i-3\hat j+4\hat k$ , giving $\vec r=(-2\hat i+\hat j-\hat k)+\lambda(-\hat i-3\hat j+4\hat k)$ .

Line 1: $\dfrac{x-1}{3}=-y,\ z=-1.$ Direction $\vec d_1=(3,-1,0)$ .

Parameterise: $x=3t+1,\ y=-t,\ z=-1$ . With $y=1\Rightarrow t=-1\Rightarrow x=-2,\ z=-1.$

Line 2: $\dfrac{x}{-2}=\dfrac{y+1}{2}=\dfrac{z+2}{1}.$ Direction $\vec d_2=(-2,2,1)$ .

Parameterise: $x=-2s,\ y=2s-1,\ z=s-2$ . With $y=1\Rightarrow s=1\Rightarrow x=-2,\ z=-1.$

Both parameterisations give the same point, so the point of intersection is $P(-2,1,-1)$ .

Direction perpendicular to both $=\vec d_1\times\vec d_2$ :

\begin{vmatrix}\hat i&\hat j&\hat k\\3&-1&0\\-2&2&1\end{vmatrix}=\hat i(-1-0)-\hat j(3-0)+\hat k(6-2)=-\hat i-3\hat j+4\hat k.

(Check: $(-1,-3,4)\cdot(3,-1,0)=0$ and $(-1,-3,4)\cdot(-2,2,1)=0$ .)

Required line through $P(-2,1,-1)$ :

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

✓Final answer

(-2,1,-1)

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

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

For the lines to be perpendicular, $\theta = 90^\circ$ , so $\cos\theta = 0$ . This forces the numerator to vanish:

$ll_1 + mm_1 + nn_1 = 0.$

Option (A) is the condition for the lines to be parallel, not perpendicular.

✓Final answer
$ll_1 + mm_1 + nn_1 = 0$
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 $\vec{n_2} = (1, 2, \lambda)$ .

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

Set equal to zero:

$2 - 8 + 3\lambda = 0 \;\Rightarrow\; 3\lambda = 6 \;\Rightarrow\; \lambda = 2.$

✓Final answer
$2$
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 $\vec n_2=(1,2,\lambda)$ .

Perpendicular planes $\Rightarrow\vec n_1\cdot\vec n_2=0$ : $2(1)+(-4)(2)+3\lambda=0$ .

$2-8+3\lambda=0\Rightarrow3\lambda=6\Rightarrow\lambda=2$ .

✓Final answer
$2$

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