Question

Find the distance between the point with position vector $negative straight i with hat on top space minus space 5 space straight j with hat on top space minus space 10 space straight k with hat on top$ and the point of intersection of the line $fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 4 end fraction space equals space fraction numerator straight z minus 2 over denominator 12 end fraction$ with the plane x – y + z = 5.

Solution

The equations of line are

fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 4 end fraction space equals space fraction numerator straight z minus 2 over denominator 12 end fraction

...(1)
Any point on this line is (3r + 2, 4r –11, 12 r + 2)
Let it lie on x – y + z = 5 ....(2)
∴ 3r + 2 – 4r + 1 + 12 r + 2 = 5 ⇒ 11r = 0 ⇒ r = 0
∴ point of intersection of line (1) and plane (2) is (2, – 1, 2)
Point with position vector

negative straight i with hat on top space minus space 5 space straight j with hat on top space minus space 10 space straight k with hat on top space is space open parentheses negative 1 comma space minus 5 comma space minus 10 close parentheses.

Let d be required distance
∴ d = Distance between (2, – 1, 2) and (–1, – 5, – 10)

equals space square root of left parenthesis negative 1 minus 2 right parenthesis squared plus left parenthesis negative 5 plus 1 right parenthesis squared plus left parenthesis negative 10 minus 2 right parenthesis squared end root space equals space square root of 9 plus 16 plus 144 end root space equals space square root of 169 space equals space 13

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