Question:
What will be the distance of $ (1, 0, 2) $ from the point of intersection of plane $ x - y + z = 16 $ and the line
$ \left(\frac{x-2}{3}\right) = \left(\frac{y+1}{4}\right) = \left(\frac{z-2}{12}\right) $ ?
What will be the distance of $ (1, 0, 2) $ from the point of intersection of plane $ x - y + z = 16 $ and the line
$ \left(\frac{x-2}{3}\right) = \left(\frac{y+1}{4}\right) = \left(\frac{z-2}{12}\right) $ ?
Updated On: Jun 14, 2022
- $ 13 $ units
- $ 17 $ units
- $ 25 $ units
- $ 19 $ units
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The Correct Option is A
Solution and Explanation
Given line is $\frac{x-2}{3}$
$=\frac{y+1}{4}$
$=\frac{z-2}{12}=r$(say) $ \dots(i)$
Any point on (i) is $P(3r+2, 4r-1, 12r+2)$
The point P lies on $x - y + z = 16$ $\dots (ii)$
$\therefore 3r+2-(4r-1)+12r+2=16$
$\Rightarrow 11 r=11$
$\Rightarrow r=1$
Thus, the line cuts the plane at $P(5, 3, 14)$.
The distance of $Q(1, 0,2)$ from $P (5,3,14)$ is given by
$PQ=\sqrt{\left(5-1\right)^{2}+\left(3-0\right)^{2}+\left(14-2\right)^{2}}$
$=\sqrt{16+9+144}$
$=13$ units
$=\frac{y+1}{4}$
$=\frac{z-2}{12}=r$(say) $ \dots(i)$
Any point on (i) is $P(3r+2, 4r-1, 12r+2)$
The point P lies on $x - y + z = 16$ $\dots (ii)$
$\therefore 3r+2-(4r-1)+12r+2=16$
$\Rightarrow 11 r=11$
$\Rightarrow r=1$
Thus, the line cuts the plane at $P(5, 3, 14)$.
The distance of $Q(1, 0,2)$ from $P (5,3,14)$ is given by
$PQ=\sqrt{\left(5-1\right)^{2}+\left(3-0\right)^{2}+\left(14-2\right)^{2}}$
$=\sqrt{16+9+144}$
$=13$ units
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Direction Cosines and Direction Ratios of Line:
Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
