Q1. The foot of the perpendicular drawn from the origin to a plane is (2, 1, 5). Find the equation of the plane.
Solution
Since, the foot of the perpendicular to the plane is A(2, 1, 5). Therefore, (2, 1, 5) is the point on the plane.
So, equation of the plane passing through the point (2,.1, 5) is:
a(x – 2) + b(y – 1) + c(z – 5) = 0.
Now, the direction ratios of the perpendicular line OA = 2 – 0, 1 – 0, 5 – 0, i.e., 2, 1, 5.
Therefore, the required plane is:
2(x – 2) + 1(y – 1) + 5(z – 5) = 0
i.e, 2x + y + 5z = 30.
Q2. Show that the planes 3x + 4y – 5z + 7 = 0 and x + 3y + 3z + 7 = 0 are perpendicular.
Solution
The direction ratios of the normals to the planes are 3, 4, – 5 and 1, 3, 3.
Now, (3)(1) + (4)(3) + (– 5)(3) = 3 + 12 – 15 = 15 – 15 = 0.
Therefore, the planes are perpendicular to each other.
Q3. Find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5).
Solution
The equation of the plane passing through the point (3, – 3, 1) is:
a(x – 3) + b(y + 3) + c(z – 1) = 0 and the direction ratios of the line joining the points
(3, 4, – 1) and (2, – 1, 5) is 2 – 3, – 1 – 4, 5 + 1, i.e., – 1, – 5, 6.
Since the plane is perpendicular to the line whose direction ratios are – 1, – 5, 6, therefore, direction ratios of the normal to the plane is – 1, – 5, 6.
So, required equation of plane is: – 1(x – 3) – 5(y + 3) + 6(z – 1) = 0
i.e., x + 5y – 6z + 18 = 0.
Q4. Show that the points A (2, 3, - 4), B (1, - 2, 3) and C (3, 8, - 11) are collinear.
Solution
Direction ratios of line joining A and B are 1 – 2, – 2 – 3, 3 + 4 i.e., – 1, – 5, 7.
The direction ratios of line joining B and C are 3 –1, 8 + 2, – 11 – 3, i.e., 2, 10, – 14.
It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel
to BC.
But point B is common to both AB and BC. Therefore, A, B, C are collinear points.
Q5. Find the equation of the plane
determined by the points A(3, -1, 2), B(5,2,4) and
C(-1,-1,6) and hence find the distance between the plane and the point
P(6,5,9).
Solution
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