# Find the coordinates of the foot of the perpendicular drawn from the point (2, 3, 7) to the plane 3x – y – z = 7. Also, find the length of the perpendicular.

Let point P = (2, 3, 7) and Q be the foot of the perpendicular drawn from P to the plane 3x – y – z = 7.

Direction ratios of PQ are proportional to 3, –1, –1 as PQ is normal to the plane.

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) = (2, 3, 7) and (l, m, n) = (3, –1, –1)

Hence, the equation of PQ is

x = 3α + 2, y = 3 – α, z = 7 – α

Let Q = (3α + 2, 3 – α, 7 – α).

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

3(3α + 2) – (3 – α) – (7 – α) = 7

9α + 6 – 3 + α – 7 + α = 7

11α – 4 = 7

11α = 11

α = 1

We have Q = (3α + 2, 3 – α, 7 – α)

Q = (3×1 + 2, 3 – 1, 7 – 1)

Q = (5, 2, 6)

Using the distance formula, we have

Thus, the required foot of perpendicular is (5, 2, 6) and the length of the perpendicular is units.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
The Embryonic Development62 mins
NCERT Exemplar Special | The Living World Part 225 mins
Nernst Equation - Learn the Concept33 mins
Revising the basics of Organic Chemistry35 mins
Understanding the Combination of cells54 mins
Nucleophilic Substitution Reaction | Getting the basics39 mins
Ray Optics | Getting the basics46 mins
Types of solution on the basis of Raoult's Law44 mins
Understant the concept of Development of pollen grains55 mins