Question 55 of 521MediumMCQ
The perpendicular distance, of the line 2x−1=−1y+2=2z+3 from the point P(2,−10,1), is :
Solutions · 1
Official solution
To find the perpendicular distance from the point P(2,−10,1) to the line given by
\[\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2},\]
follow these steps:
Parametrize the Line:
From the given symmetric equations, set the common parameter as t:
x=1+2t
y=−2−t
z=−3+2t
This shows that:
A point on the line is A(1,−2,−3) (when t=0).
The direction vector is \[\vec{d} = \langle 2, -1, 2 \rangle.\]
Determine the Vector from Point A to P:
Calculate
\[\vec{AP} = P - A = \langle 2 - 1, \; -10 - (-2), \; 1 - (-3) \rangle = \langle 1, \; -8, \; 4 \rangle.\]
Compute the Cross Product:
The formula for the perpendicular distance from a point to a line in 3D is:
\[d = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}.\]
First, find the cross product \[\vec{AP} \times \vec{d}\]:
\[ \vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix} = \Big( (-8)(2) - (4)(-1), \; (4)(2) - (1)(2), \; (1)(-1) - (-8)(2) \Big). \]
Evaluate each component:
First component: \[(-8 \times 2) - (4 \times -1) = -16 + 4 = -12.\]
Second component: \[ (4 \times 2) - (1 \times 2) = 8 - 2 = 6.\]
Third component: \[ (1 \times -1) - (-8 \times 2) = -1 + 16 = 15.\]
So,
\[\vec{AP} \times \vec{d} = \langle -12, 6, 15 \rangle.\]
Calculate the Magnitudes:
For the cross product:
\[ \|\vec{AP} \times \vec{d}\| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405} = 9\sqrt{5}. \]
For the direction vector:
\[ \|\vec{d}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. \]
Compute the Distance:
Substitute the magnitudes into the distance formula:
\[ d = \frac{9\sqrt{5}}{3} = 3\sqrt{5}. \]
Thus, the perpendicular distance from the point P(2,−10,1) to the line is \[3\sqrt{5}\].
Comparing with the options given, the correct answer is:
Option C: \[3\sqrt{5}\].
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