Question

Find the distance of the point with position vector -i - 5j - 10k from the point of intersection of the line r = (2i - j + 2k) + λ (3i + 4j + 12k) with the plane \(\vec r(\hat i-\hat j+\hat k)=5.\)

Answer 1

Let P be the point with position vector \(\vec p=-\hat i-5\hat j-10\hat k\) and Q be the point of intersection of the given line and the plane.

We have the line equation as

Let the position vector of Q be \(\vec q\). As Q is a point on this line, for some scalar α, we have

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

\(\vec r.(\hat i-\hat j+\hat k)=5.\)

⇒ (2 + 3α)(1) + (–1 + 4α)(–1) + (2 + 12α)(1) = 5

⇒ 2 + 3α + 1 – 4α + 2 + 12α = 5

⇒ 11α + 5 = 5

⇒ 11α = 0 

∴ α = 0

We have

Using the distance formula, we have

Thus, the required distance is 13 units.