Question:
If $(a, b, c)$ are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$, then the point $P = (a, 3b, 2c)$ lies on the plane
If $(a, b, c)$ are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$, then the point $P = (a, 3b, 2c)$ lies on the plane
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Direction ratios = difference of coordinates; check which plane the derived point satisfies.
Updated On: May 18, 2025
- $x + y + z = 0$
- $x + y - 2z = 0$
- $x + 2y + 3z = 0$
- $x - 2y + 3z = 0$
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The Correct Option is B
Solution and Explanation
Direction ratios of line joining $(4,3,-5)$ and $(-2,1,-8)$ are:
$a = -6,\ b = -2,\ c = -3$
Then $P = (-6, -6, -6)$
Now check which plane this satisfies.
Substitute into each:
Only $x + y - 2z = -6 -6 -2(-6) = -12 +12 = 0$
So, this is the correct plane.
$a = -6,\ b = -2,\ c = -3$
Then $P = (-6, -6, -6)$
Now check which plane this satisfies.
Substitute into each:
Only $x + y - 2z = -6 -6 -2(-6) = -12 +12 = 0$
So, this is the correct plane.
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