"The square of the distance of the point of intersection of the line ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$ and the plane $2x - y + z = 6$ from the point ($-$1, $-$1, 2) is __________."

Question

Accepted Answer

"${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6} = \lambda$

$x = 2\lambda + 1,y = 3\lambda + 2,z = 6\lambda - 1$

for point of intersection of line & plane

$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$

$7\lambda = 7 \Rightarrow \lambda = 1$

point : (3, 5, 5)

(distance)² = ${(3 + 1)^2} + {(5 + 1)^2} + {(5 - 2)^2}$

$= 16 + 36 + 9 = 61$"

The square of the distance of the point of intersection

of the line ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$ and the plane $2x - y + z = 6$ from the point ($-$1, $-$1, 2) is __________.

Solution

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The square of the distance of the point of intersection of the line ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$ and the plane $2x - y + z = 6$ from the point ($-$1, $-$1, 2) is __________.

Solution

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More Three Dimensional Geometry questions

Drill 25 more like these. Every day. Free.

The square of the distance of the point of intersection

of the line ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$ and the plane $2x - y + z = 6$ from the point ($-$1, $-$1, 2) is __________.