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प्रश्न
The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is ______
पर्याय
x - 3y - 2z = -2
2x - z = 2
x - y - z = 0
x + 3y + z = 4
MCQ
रिकाम्या जागा भरा
उत्तर
The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is x - y - z = 0.
Explanation:
The equation of the required plane is given by
2x - y - 4 + λ(y + 2z - 4) = 0 ..................(i)
This plane passes through (1, 1, 0)
∴ 2(1) - 1 - 4 + λ(1 + 0 - 4) = 0
⇒ λ = -1
Substitution λ = -1 in (i), we get
2x - y - 4 - 1(y + 2z - 4) = 0
⇒ x - y - z = 0
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