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CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q27: Plane through (1,0,0)(1, 0, 0) containing the line r⃗=λj^\vec r = \lambda\hat j

Sample 2020-21Ch 11 · Three Dimensional GeometryCh 10 · Vector Algebra
The question

Find the vector equation of the plane that passes through the point (1,0,0) and contains the line r⃗=λj^\vec r = \lambda\hat j⁠.

This question uses the equation of a plane. This topic is no longer in the CBSE syllabus (2025-26).

Exam answer

Let n⃗\vec n be a normal to the plane. The plane passes through (1,0,0)(1, 0, 0)⁠, whose position vector is i^\hat i⁠, so its equation is

(r⃗−i^)⋅n⃗=0…(i)(\vec r - \hat i)\cdot\vec n = 0 \quad\ldots(\text{i})
Why?

A plane through the point with position vector a⃗\vec a⁠, with normal n⃗\vec n⁠, has the equation

(r⃗−a⃗)⋅n⃗=0(\vec r - \vec a)\cdot\vec n = 0

The point (1,0,0)(1, 0, 0) has position vector 1i^+0j^+0k^=i^1\hat i + 0\hat j + 0\hat k = \hat i⁠. The normal n⃗\vec n is not known yet; the next steps find it.

The plane contains the line r⃗=0⃗+λj^\vec r = \vec 0 + \lambda\hat j⁠. So O lies in it (putting r⃗=0⃗\vec r = \vec 0 in (i)), and it is parallel to j^\hat j⁠.

∴ i^⋅n⃗=0 and j^⋅n⃗=0\therefore\ \hat i\cdot\vec n = 0 \text{ and } \hat j\cdot\vec n = 0
Why?

The line passes through O (at λ=0\lambda = 0⁠), so O is on the plane. Putting r⃗=0⃗\vec r = \vec 0 in (i) gives −i^⋅n⃗=0-\hat i\cdot\vec n = 0⁠, that is i^⋅n⃗=0\hat i\cdot\vec n = 0⁠.

(0⃗−i^)⋅n⃗=0−i^⋅n⃗=0i^⋅n⃗=0(×−1)\begin{aligned} (\vec 0 - \hat i)\cdot\vec n &= 0 \\ -\hat i\cdot\vec n &= 0 \\ \hat i\cdot\vec n &= 0 \quad (\times -1) \end{aligned}

The line lies in the plane, so its direction j^\hat j is perpendicular to the normal: j^⋅n⃗=0\hat j\cdot\vec n = 0⁠.

⇒n⃗=k^\Rightarrow \vec n = \hat k⁠. Hence the equation of the plane is

Why?

n⃗\vec n is perpendicular to both i^\hat i and j^\hat j⁠, so it is along their cross product:

i^×j^=k^\hat i \times \hat j = \hat k

Any non-zero multiple of k^\hat k would give the same plane.

(r⃗−i^)⋅k^=0(\vec r - \hat i)\cdot\hat k = 0 ⇒r⃗⋅k^−i^⋅k^=0\Rightarrow \vec r\cdot\hat k - \hat i\cdot\hat k = 0
Why?

i^\hat i and k^\hat k are perpendicular, so i^⋅k^=0\hat i\cdot\hat k = 0⁠.

r⃗⋅k^−i^⋅k^=0r⃗⋅k^−0=0r⃗⋅k^=0\begin{aligned} \vec r\cdot\hat k - \hat i\cdot\hat k &= 0 \\ \vec r\cdot\hat k - 0 &= 0 \\ \vec r\cdot\hat k &= 0 \end{aligned}

With r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠, the answer says z=0z = 0⁠: the xy-plane, which does contain (1,0,0)(1, 0, 0) and the y-axis.

∴\therefore the plane is r⃗⋅k^=0\vec r\cdot\hat k = 0

Where this comes from

Chapter 11, Three Dimensional Geometry: Vector equation of a plane containing a line.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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Three Dimensional Geometry in past papers: question 34 of 35 · see them all