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
(a)
2x − z = 2
(b)
x − 3y − 2z = −2
(c)
x − y − z = 0
(d)
x + 3y + z = 4
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
(a)
2x − z = 2
(b)
x − 3y − 2z = −2
(c)
x − y − z = 0
(d)
x + 3y + z = 4
Question ID - 57566 | SaraNextGen Top Answer
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
(a)
2x − z = 2
(b)
x − 3y − 2z = −2
(c)
x − y − z = 0
(d)
x + 3y + z = 4
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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 |
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(a) |
2x − z = 2 |
(b) |
x − 3y − 2z = −2 |
|
(c) |
x − y − z = 0 |
(d) |
x + 3y + z = 4 |
1 Answer
127 votes
Answer Key / Explanation : (c) -
Let the equation of required plane be;
(2x − y − 4) + 
(y + 2z − 4) = 0
This plane passes through (1, 1, 0) then
(2 − 1 − 4) + 
(1 + 0 − 4) = 0
⇒ 
= − 1
Equation of required plane will be
(2x − y − 4) − (y + 2z − 4) = 0
⇒ 2x − 2y − 2z = 0
⇒ x − y − z = 0
127 votes
Answer Link
127