NIMCET 2016 Mathematics PYQ — The vector perpendicular to the plane passing through (1, -1, 0) … | Mathem Solvex | Mathem Solvex
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NIMCET 2016 — Mathematics PYQ
NIMCET | Mathematics | 2016
The vector perpendicular to the plane passing through (1, -1, 0) (2, 1, -2) and (-1, 1, 2) is
Choose the correct answer:
A.
6j+6k
(Correct Answer)
B.
6i+7k
C.
7i+6k
D.
7i+8k
Correct Answer:
6j+6k
Explanation
Concept: To find a vector perpendicular to the find two plane, first find the vectors in the plane and then take their cross product.
Calculations: Given, the plane is passing through the point (1, -1, 0) (2, 1, -2) and (-1, 1, 2) Step 1) Find two vectors in the plane. We will do this by finding the vector from (1, -1, 0) (2, 1, -2) and from(1, -1, 0) (-1, 1, 2). As all three points are in the plane, so will each of those vectors.
u1=(1,−1,0)−(2,1,−2)=(−1,−2,2)
u2=(1,−1,0)−(−1,1,2)=(2,−2,2)
Step 2) Find a vector perpendicular to the plane.
If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces a vector perpendicular to both, we will use the cross product of u1 and u2 to find a vector u perpendicular to the plane containing them.
⇒u=u1×u2
⇒u=i−12amp;jamp;−2amp;−2amp;kamp;2amp;2
⇒u=i(−4+4)−j(−2−4)+k(2+4)
⇒u=6j+6k
Hence, the vector perpendicular to the plane passing through (1,−1,0)(2,1,−2) and (−1,1,2) is u=6j+6k.
Explanation
Concept: To find a vector perpendicular to the find two plane, first find the vectors in the plane and then take their cross product.
Calculations: Given, the plane is passing through the point (1, -1, 0) (2, 1, -2) and (-1, 1, 2) Step 1) Find two vectors in the plane. We will do this by finding the vector from (1, -1, 0) (2, 1, -2) and from(1, -1, 0) (-1, 1, 2). As all three points are in the plane, so will each of those vectors.
u1=(1,−1,0)−(2,1,−2)=(−1,−2,2)
u2=(1,−1,0)−(−1,1,2)=(2,−2,2)
Step 2) Find a vector perpendicular to the plane.
If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces a vector perpendicular to both, we will use the cross product of u1 and u2 to find a vector u perpendicular to the plane containing them.
⇒u=u1×u2
⇒u=i−12amp;jamp;−2amp;−2amp;kamp;2amp;2
⇒u=i(−4+4)−j(−2−4)+k(2+4)
⇒u=6j+6k
Hence, the vector perpendicular to the plane passing through (1,−1,0)(2,1,−2) and (−1,1,2) is u=6j+6k.
