CBSE Sample Paper 2020-21 · Part A: Section I · 1 mark

Q14: Where the line $\frac{x + 3}{3} = \frac{y - 1}{-1} = \frac{z - 5}{-5}$ cuts the XY plane

The question

Find the coordinates of the point where the line $\frac{x + 3}{3} = \frac{y - 1}{-1} = \frac{z - 5}{-5}$ cuts the XY plane.

Exam answer

Let $\frac{x + 3}{3} = \frac{y - 1}{-1} = \frac{z - 5}{-5} = \lambda$ ⁠.

Any point on the line is

(3\lambda - 3,\ -\lambda + 1,\ -5\lambda + 5)

Why?

Calling the common value $\lambda$ lets us write every point of the line with one letter. Solve each equal ratio for its coordinate:

\begin{aligned} x + 3 = 3\lambda &\Rightarrow x = 3\lambda - 3 \\ y - 1 = -\lambda &\Rightarrow y = -\lambda + 1 \\ z - 5 = -5\lambda &\Rightarrow z = -5\lambda + 5 \end{aligned}

On the XY plane, $z = 0$ ⁠:

-5\lambda + 5 = 0 \Rightarrow \lambda = 1

Why?

Every point of the XY plane has $z$ -coordinate 0, so we set the $z$ -coordinate of the general point equal to 0 and solve for $\lambda$ ⁠.

\begin{aligned} -5\lambda + 5 &= 0 \\ -5\lambda &= -5 \\ \lambda &= \frac{-5}{-5} = 1 \end{aligned}

The point is $(3 - 3,\ -1 + 1,\ -5 + 5)$ ⁠.

Why?

Put $\lambda = 1$ into each coordinate:

\begin{aligned} x &= 3(1) - 3 = 3 - 3 = 0 \\ y &= -(1) + 1 = -1 + 1 = 0 \\ z &= -5(1) + 5 = -5 + 5 = 0 \end{aligned}

Check with the line: $\frac{0 + 3}{3}$ ⁠, $\frac{0 - 1}{-1}$ and $\frac{0 - 5}{-5}$ all equal 1, so the origin is on the line.

$\therefore$ the line cuts the XY plane at $(0, 0, 0)$ ⁠.

Where this comes from

Chapter 11, Three Dimensional Geometry: Point where a line meets the XY-plane.

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Three Dimensional Geometry in past papers: question 33 of 35 · see them all

Q14: Where the line x+33=y−1−1=z−5−5\frac{x + 3}{3} = \frac{y - 1}{-1} = \frac{z - 5}{-5}3x+3​=−1y−1​=−5z−5​ cuts the XY plane