The coordinates of the point where the line f | vedclass

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(D) Let the point on the line be $(x, y, z)$.Given the line equation: $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4} = r$.Expressing coordinate

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$(5, -1, 1)$

Vedclass · Answer

(D) Let the point on the line be $(x, y, z)$.Given the line equation: $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4} = r$.Expressing coordinates in terms of $r$: $x = -r + 6$,$y = -1$,$z = 4r - 3$.Since the point lies on the plane $x + y - z = 3$,substitute these values into the plane equation:$(-r + 6) + (-1) - (4r - 3) = 3$.$-r + 6 - 1 - 4r + 3 = 3$.$-5r + 8 = 3$.$-5r = -5$.$r = 1$.Substituting $r = 1$ back into the coordinate expressions:$x = -1 + 6 = 5$.$y = -1$.$z = 4(1) - 3 = 1$.Thus,the required coordinates are $(5, -1, 1)$.

The coordinates of the point where the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ meets the plane $x + y - z = 3$ are

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