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Question

Find the image of the point with position vector 3i^+j^+2k^ in the plane r·2i^-j^+k^=4. Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through 3i^+j^+2k^.

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Solution


Let Q be the image of the point P (3 i^+j^+2 k^) in the plane r. 2 i^-j^+ k^ = 4Since PQ passes through P and is normal to the given plane, it is parallel to the normal vector 2 i^-j^+k^. So, the equation of PQ isr=3 i^+j^+2 k^+λ 2 i^-j^+k^As Q lies on PQ, let the position vector of Q be 3+2λ i^+1-λ j^+2+λ k^ .Let R be the mid-point of PQ. Then, the position vector of R is 3+2λ i^+1-λ j^+2+λ k^+3 i^+j^+2 k^2=6+2λ i^+2-λ j^+4+λ k^2=3+λ i^+1-λ2 j^+2+λ2 k^Since R lies in the plane r. 2 i^-j^+ k^ = 4,3+λ i^+1-λ2 j^+2+λ2 k^. 2 i^-j^+ k^ = 46+2λ-1+λ2+2+λ2=47 + 2λ + λ2 + λ2 = 4 14 + 6 λ = 8 6 λ = 8 - 14 λ = -1Putting λ = -1 in Q, we get Q= 3+2(-1) i^+1-(-1) j^+2+(-1) k^ = i^ + 2 j^ +k ^or (1, 2, 1)Therefore, by putting λ = -1 in R, we get R = 3+(-1) i^+1-(-1)2 j^+2+(-1)2 k^= 2 i^ + 32 j^ +32 k^

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