Twice the distance of the point (1, 0, 2) from the point of intersection of the line and the plane is

Question

Twice the distance of the point (1, 0, 2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

Smart Achievers · Accepted Answer

(4)

$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = λ [say]$ $. . . (i)$

$And equation of plane is x - y + z = 16$

$Any point on the line (i) is (3 λ + 2, 4 λ - 1, 12 λ + 2)$

$Let this point of intersection of the line and plane.$

$(3 λ + 2) - (4 λ - 1) + (12 λ + 2) = 16$

$∴$ $11 λ = 11$ $\Rightarrow λ = 1$

$So, the point of intersection is (5, 3, 14)$

$Now, distance between the points (1, 0, 2) and (5, 3, 14)$

$= \sqrt{{(5 - 1)}^{2} + {(3 - 0)}^{2} + {(14 - 2)}^{2}}$

$= \sqrt{16 + 9 + 144}$

$= \sqrt{169}$

$= 13$

$∴$ $Twice the distance = 26$

Solution

Q.
Twice the distance of the point (1, 0, 2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

1 $2 \sqrt{14}$

2 $8$

3 $3 \sqrt{21}$

4 $26$

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Solution

Q. Twice the distance of the point (1, 0, 2) from the point of intersection of the line x-23=y+14=z-212 and the plane x-y+z=16 is

1 214

2 8

3 321