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If the planes 2x−3y+4z+5=0, x+2y–kz+7=0 are perpendicular then k=

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a

4

b

1

c

6

d

-1

answer is D.

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Detailed Solution

2x–3y+4z+5=0x+2y–kz+7 =0 are ⊥ ler planes thena1a2+b1b2+c1c2+0 2–6–4 k=0 4k=− 4 k=−1

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