Question 89 of 521MediumMCQ
The plane, passing through the points (0,−1,2) and (−1,2,1) and parallel to the line passing through (5,1,−7) and (1,−1,−1), also passes through the point :
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Official solution
The first step is to find the normal vector to the desired plane. Since the plane is parallel to the line passing through the points (5, 1, -7) and (1, -1, -1), the direction vector of that line is also parallel to the plane. The direction vector is the difference between the coordinates of the two points, which is (5-1, 1-(-1), -7-(-1)) = (4, 2, -6).
Next, let's find another vector that is parallel to the plane. This vector can be obtained by taking the difference between the coordinates of the points (0, -1, 2) and (-1, 2, 1), which the plane passes through. This gives us a vector of (0-(-1), -1-2, 2-1) = (1, -3, 1).
The normal to the plane is perpendicular to both these vectors. It can be found by taking the cross product of the two vectors.
The cross product of vectors (4, 2, -6) and (1, -3, 1) is :
n=∣i^j^k^42−61−31∣=i^(−16)−j^(+10)+k^(−14)=−16i^−10j^−14k^
The equation of the plane can now be written in the form :
-16x - 10y - 14z = d
We can find the constant 'd' by substituting one of the points through which the plane passes, say (0, -1, 2) :
d = -16.0 -10.(-1) -14.2 = 10 - 28 = -18
So, the equation of the plane is -16x - 10y - 14z = -18.
Now, we substitute the given options into the equation to check which one satisfies it :
Option A: (-16.0 -10.5 -14.(-2) = -18) => -50 + 28 = -22 \neq -18, so A is not correct.
Option B: (-16.2 -10.0 -14.1 = -18) => -32 - 14 = -46 \neq -18, so B is not correct.
Option C: (-16.1 -10.(-2) -14.1 = -18) => -16 + 20 -14 = -10 \neq -18, so C is not correct.
Option D: (-16.(-2) -10.5 -14.0 = -18) => 32 - 50 = -18, so D is correct.
So, the plane also passes through the point given in Option D, which is (-2, 5, 0).
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