Find the distance of the point (1, -2, 3) from the plane x – y + z = 5 measured along a line parallel to x/2 = y/3 = z/-6.

Question

Find the distance of the point (1, -2, 3) from the plane x – y + z = 5 measured along a line parallel to \(\cfrac{\text x}2=\cfrac{y}3=\cfrac{z}{-6}\).

Answer 1

Let point P = (1, –2, 3).

We need to find distance from P to the plane x – y + z = 5 measured along a line parallel to \(\cfrac{\text x}2=\cfrac{y}3=\cfrac{z}{-6}\).

Let the line drawn from P parallel to the given line meet the plane at Q.

Direction ratios of PQ are proportional to 2, 3, –6 as PQ is parallel to the given line.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (0, 0, 0) and (l, m, n) = (2, 3, –6)

Hence, the equation of PQ is

⇒ x = 2α + 1, y = 3α – 2, z = –6α + 3

Let Q = (2α + 1, 3α – 2, –6α + 3).

This point lies on the given plane, which means this point satisfies the plane equation.

⇒ (2α + 1) – (3α – 2) + (–6α + 3) = 5

⇒ 2α + 1 – 3α + 2 – 6α + 3 = 5

⇒ –7α + 6 = 5

⇒ –7α = –1

\(\therefore\alpha=\cfrac17\) 

We have Q = (2α + 1, 3α – 2, –6α + 3)

Using the distance formula, we have

Thus, the required distance is 1 unit.