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Question: The direction cosines of the projection of the line $\frac{1}{2}(x - 1) = -y = z + 2$ on the plane $...

The direction cosines of the projection of the line 12(x−1)=−y=z+2\frac{1}{2}(x - 1) = -y = z + 2 on the plane 2x+y−3z=42x + y - 3z = 4 are-

A

(26,16,16)(\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})

B

(26,16,16)(\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})

C

(26,16,16)(\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})

D

None of these

Answer

None of these

Explanation

Solution

The given line is 12(x−1)=−y=z+2\frac{1}{2}(x - 1) = -y = z + 2. We can write this in the standard symmetric form as x−12=y−0−1=z−(−2)1\frac{x - 1}{2} = \frac{y - 0}{-1} = \frac{z - (-2)}{1}.
The direction vector of the line is v⃗=(2,−1,1)\vec{v} = (2, -1, 1). The line passes through the point P0(1,0,−2)P_0(1, 0, -2).

The given plane is 2x+y−3z=42x + y - 3z = 4. The normal vector to the plane is n⃗=(2,1,−3)\vec{n} = (2, 1, -3).

To find the direction of the projection of the line on the plane, we first check if the line is parallel to the plane. The line is parallel to the plane if its direction vector is perpendicular to the normal vector of the plane. We calculate the dot product of v⃗\vec{v} and n⃗\vec{n}:

v⃗⋅n⃗=(2)(2)+(−1)(1)+(1)(−3)=4−1−3=0\vec{v} \cdot \vec{n} = (2)(2) + (-1)(1) + (1)(-3) = 4 - 1 - 3 = 0.

Since v⃗⋅n⃗=0\vec{v} \cdot \vec{n} = 0, the direction vector of the line is perpendicular to the normal vector of the plane. This means the line is parallel to the plane.

Next, we check if the line lies in the plane. The line lies in the plane if a point on the line also lies in the plane. The point P0(1,0,−2)P_0(1, 0, -2) lies on the line. We substitute the coordinates of P0P_0 into the equation of the plane:

2(1)+(0)−3(−2)=2+0+6=82(1) + (0) - 3(-2) = 2 + 0 + 6 = 8.

The equation of the plane is 2x+y−3z=42x + y - 3z = 4. Since 8≠48 \neq 4, the point P0P_0 does not lie in the plane.

Since the line is parallel to the plane but does not lie in the plane, the projection of the line on the plane is a line parallel to the original line. Therefore, the direction vector of the projected line is the same as the direction vector of the original line, or its negative. The direction vector of the original line is v⃗=(2,−1,1)\vec{v} = (2, -1, 1).

The direction cosines of the projected line are the components of the unit vector in the direction of v⃗\vec{v}. The magnitude of v⃗\vec{v} is ∥v⃗∥=22+(−1)2+12=4+1+1=6\|\vec{v}\| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}. The unit vector in the direction of v⃗\vec{v} is v^=v⃗∥v⃗∥=(2,−1,1)6=(26,−16,16)\hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{(2, -1, 1)}{\sqrt{6}} = (\frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}).

The direction cosines are (26,−16,16)(\frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}). Another possible set of direction cosines is the negative of these values: (−26,16,−16)(\frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-1}{\sqrt{6}}).

We compare these direction cosines with the given options. The options are (26,16,16)(\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}). This does not match our calculated direction cosines.

Based on the given line and plane equations, the line is parallel to the plane and does not lie in it. The projection is a parallel line with direction cosines (26,−16,16)(\frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}) or (−26,16,−16)(\frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-1}{\sqrt{6}}). None of the options match this result. Therefore, the correct option is (D) None of these.