Question

find the value of k so that the plane (k+1)x-y+(2-k)z=5 is perpendicular to the plane 2x+6y-z+3=0​

Answer 1

Answer:

$0+ 1 \times 10 - 9 + 2 - 0 \times z = 5 \\ 0+ 1 = 1 + 2 = 3 - 0 = 3 \times 5 \\ 2 \times 10 = 20 + 6 \times 0 = 0 - 10 + 3 = 0 \\ 13 \: ans....$

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Answer 2

If the planes are ⊥ , then the dot product of directionratios is 0⇒(3,−6,−2).(2,1,−K)=0⇒6−6+2K=0⇒K=0