Q. 74.0( 4 Votes )

Find the coordina

Answer :

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.


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