(b) Find the vector and the cartesian equation of the line that passes through (−1, 2, 7) and is perpendicular

Question

Q. (b) Find the vector and the cartesian equation of the line that passes through (-1, 2, 7) and is perpendicular to the lines 𝑟⃗ = 2ı̂ + ȷ̂ - 3k̂ + λ(ı̂ + 2ȷ̂ + 5k̂ ) and 𝑟⃗ = 3ı̂ + 3ȷ̂ - 7k̂ + μ(3ı̂ - 2ȷ̂ + 5k̂ ).

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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

\vec{r} = \vec{a} + \lambda\,(5\hat{i} + 2\hat{j} - 3\hat{k}).

Tip

If $\vec{b_1}\times\vec{b_2} = \vec{0}$ , the two given lines are parallel, so infinitely many perpendicular directions exist — the cross product cannot single one out.

The Takeaway

Whenever you must build a line perpendicular to two given lines, reach for the cross product of their direction vectors. It hands you the perpendicular direction in one step, and you attach the required point to complete the line's equation.

Using the cross product to find a line perpendicular to two given lines is a standard application drawn from the NCERT Class 12 Vector Algebra and Three Dimensional Geometry chapters, tested in CBSE boards, JEE Main and several state CETs. "How to find a vector perpendicular to two vectors" is a commonly searched query, and this same technique underlies finding the normal to a plane containing two given directions.

A line perpendicular to two given lines must be parallel to the cross product of their direction vectors. Using the point (−1,2,7) and the cross product direction, we obtain the vector equation $\vec{r} = (-\hat{i}+2\hat{j}+7\hat{k}) + t(2\hat{i}+\hat{j}-\hat{k})$ and the cartesian equation $\frac{x+1}{2} = \frac{y-2}{1} = \frac{z-7}{-1}$ .

Concept and Intuition

When a line is perpendicular to two other lines, it means its direction vector is perpendicular to both of their direction vectors. In 3D geometry, the vector that is perpendicular to two given vectors is their cross product. So the direction of the required line is simply the cross product of the direction vectors of the two given lines.

This is a standard and powerful technique: instead of solving simultaneous equations, we directly compute the perpendicular direction.

Tip

Always check that the two given direction vectors are not parallel — otherwise their cross product would be zero and no unique perpendicular line exists. Here, $(1,2,5)$ and $(3,-2,5)$ are clearly not scalar multiples, so we are safe.

Step-by-step solution

1. Identify the direction vectors of the given lines.

The first line is $\vec{r} = (2\hat{i}+\hat{j}-3\hat{k}) + \lambda(\hat{i}+2\hat{j}+5\hat{k})$ , so its direction vector is $\vec{d}_1 = \hat{i}+2\hat{j}+5\hat{k}$ .

The second line is $\vec{r} = (3\hat{i}+3\hat{j}-7\hat{k}) + \mu(3\hat{i}-2\hat{j}+5\hat{k})$ , so its direction vector is $\vec{d}_2 = 3\hat{i}-2\hat{j}+5\hat{k}$ .

2. Find a vector perpendicular to both $\vec{d}_1$ and $\vec{d}_2$ .

The required direction $\vec{d}$ is the cross product $\vec{d}_1 \times \vec{d}_2$ .

Compute using the determinant:

\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 5 \\ 3 & -2 & 5 \end{vmatrix}

Expanding:

\vec{d} = \hat{i}(2\cdot5 - 5\cdot(-2)) - \hat{j}(1\cdot5 - 5\cdot3) + \hat{k}(1\cdot(-2) - 2\cdot3)

= \hat{i}(10 + 10) - \hat{j}(5 - 15) + \hat{k}(-2 - 6)

= 20\hat{i} - (-10)\hat{j} + (-8)\hat{k}

= 20\hat{i} + 10\hat{j} - 8\hat{k}

We can simplify by dividing by the common factor 2:

\vec{d} = 10\hat{i} + 5\hat{j} - 4\hat{k} \quad \text{or simply} \quad \vec{d} = 2\hat{i} + \hat{j} - \hat{k}

Note

Any scalar multiple of a direction vector represents the same line direction. Using the simplest integer form $2\hat{i}+\hat{j}-\hat{k}$ makes the cartesian equation cleaner.

3. Write the vector equation of the required line.

The line passes through point $P(-1,2,7)$ , whose position vector is $\vec{a} = -\hat{i}+2\hat{j}+7\hat{k}$ .

Using direction $\vec{d} = 2\hat{i}+\hat{j}-\hat{k}$ , the vector equation is:

\vec{r} = \vec{a} + t\vec{d}

\boxed{\vec{r} = (-\hat{i}+2\hat{j}+7\hat{k}) + t(2\hat{i}+\hat{j}-\hat{k})}

4. Convert to cartesian equation.

If $\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$ , then equating components:

x = -1 + 2t, \quad y = 2 + t, \quad z = 7 - t

Solving for $t$ from each:

t = \frac{x+1}{2}, \quad t = \frac{y-2}{1}, \quad t = \frac{z-7}{-1}

Since $t$ is the same parameter, we equate:

\boxed{\frac{x+1}{2} = \frac{y-2}{1} = \frac{z-7}{-1}}

Watch out

A common mistake is to forget the negative sign in the denominator for $z$ . The direction vector component for $z$ is $-1$ , so the denominator is $-1$ , not $1$ . Always check the sign of each component.

✓Final answer

\vec{r} = (-\hat{i}+2\hat{j}+7\hat{k}) + t(2\hat{i}+\hat{j}-\hat{k})

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

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

Direction is $\hat{k}$ , which is fine, but the point used is $(\hat{i} - 5\hat{j} - 4\hat{k})$ , i.e., $(1, -5, -4)$ . That’s not the given point $(-1, 5, 4)$ . So this is wrong.

Option (D): $\vec{r} = \lambda\hat{k}$

This passes through the origin $(0,0,0)$ , not through $(-1,5,4)$ . Wrong.

Watch out

A common mistake is to pick option (C) because it has $\lambda\hat{k}$ , forgetting that the point must match exactly. The direction alone isn’t enough — the line must pass through the specified point.

Tip

When a plane equation is $z = 0$ , its normal vector is $\hat{k}$ . A line perpendicular to the plane is parallel to this normal. So the direction vector is simply $\hat{k}$ (or any non-zero scalar multiple).

✓Final answer
$The correct option is (B) .$
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

So the direction ratios (and hence, after normalising, direction cosines) of the required line are m₁n₂−n₁m₂, n₁l₂−l₁n₂, l₁m₂−l₂m₁.

✓Final answer
$m₁n₂ - n₁m₂, n₁l₂ - l₁n₂, l₁m₂ - l₂m₁ — option (c).$

Q.(b) Find the vector and the cartesian equation of the line that passes through (−1, 2, 7) and is perpendicular to the lines 𝑟⃗ = 2ı̂ + ȷ̂ − 3k̂ + λ(ı̂ + 2ȷ̂ + 5k̂ ) and 𝑟⃗ = 3ı̂ + 3ȷ̂ − 7k̂ + μ(3ı̂ − 2ȷ̂ + 5k̂ ).

Concept and Intuition

Step-by-step solution

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