The plane passing through the point (5, 1, 2) perpendicular to the line 2(x - 2) = y - 4 = Z-5 will meet
the line in the point
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The equation of plane passing through the point (1,1,1) and perpendicular to the planes 2x+y−2z=5 and 3x−6y−2z=7 is
December 27, 2019avatar
Vishnu Bankar
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∣
∣
∣
∣
∣
∣
∣
∣
x−1
2
3
y−1
1
−6
z−1
−2
−2
∣
∣
∣
∣
∣
∣
∣
∣
=0
⇒(x−1)(−14)−(y−1)(2)+(z−1)(−15)=0
⇒14x−14+2y−2+15z−15=0
14x+2y+15z=31
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