If $L_1$ is the line of intersection of the planes $2x - 2y + 3z - 2 =0, x - y + z + 1 = 0$ and $L_2$ is the line of intersection of the planes $x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0,$ then the distance of the origin from the plane, containing the lines $L_1$ and $L_2$, is:

Question

If the distances of the point $ (1,2,a) $ from the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] along the lines \[ L_1:\ \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} \quad \text{and} \quad L_2:\ \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} \] are equal, then $ a+b+c $ is equal to:

Answer 1

To find the distance of the origin from the plane containing the lines $L_1$ and $L_2$, we need to perform the following steps:

First, determine the direction vectors of the lines $L_1$ and $L_2$. The direction vector of the line of intersection of two planes is given by the cross product of the normals of those planes.
The normal of the first plane $2x - 2y + 3z - 2 = 0$ is $ \langle 2, -2, 3 \rangle $, and for the plane $x - y + z + 1 = 0$ is $ \langle 1, -1, 1 \rangle $. Therefore, the direction vector of $L_1$ is:
$$ \mathbf{d_1} = \langle 2, -2, 3 \rangle \times \langle 1, -1, 1 \rangle = \langle 1, -1, 0 \rangle $$
Similarly, for the line $L_2$, the normal of the plane $x + 2y - z - 3 = 0$ is $ \langle 1, 2, -1 \rangle $ and for $3x - y + 2z - 1 = 0$ is $ \langle 3, -1, 2 \rangle $. Therefore, the direction vector of $L_2$ is:
$$ \mathbf{d_2} = \langle 1, 2, -1 \rangle \times \langle 3, -1, 2 \rangle = \langle 3, -5, -7 \rangle $$
The normal to the plane containing these two lines can be found as the cross product of $ \mathbf{d_1} $ and $ \mathbf{d_2} $:
$$ \mathbf{n} = \langle 1, -1, 0 \rangle \times \langle 3, -5, -7 \rangle = \langle 7, 7, 2 \rangle $$
The equation of the plane containing the line $L_1$ is obtained by substituting a point on $L_1$. Assume $(3, 4, 7)$ is a point on $L_1$. The equation of the plane is:
$$ 7(x - 3) + 7(y - 4) + 2(z - 7) = 0 $$ which simplifies to $$7x + 7y + 2z = 67$$
Now, calculate the distance from the origin $(0,0,0)$ to this plane using the formula:
$$ D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} $$
Substitute $a = 7, b = 7, c = 2, d = -67$:
$$ D = \frac{|0 + 0 + 0 - 67|}{\sqrt{7^2 + 7^2 + 2^2}} = \frac{67}{\sqrt{49 + 49 + 4}} = \frac{67}{\sqrt{102}} = \frac{67}{3\sqrt{2}} $$
Simplified, the distance $$ D = \frac{1}{3\sqrt{2}} $$.

Therefore, the distance of the origin from the plane containing the lines $L_1$ and $L_2$ is $$ \frac{1}{3\sqrt{2}} $$.

Answer 2

To find the distance of the origin from the plane containing the lines $L_1$ and $L_2$, we need to perform the following steps:

First, determine the direction vectors of the lines $L_1$ and $L_2$. The direction vector of the line of intersection of two planes is given by the cross product of the normals of those planes.
The normal of the first plane $2x - 2y + 3z - 2 = 0$ is $ \langle 2, -2, 3 \rangle $, and for the plane $x - y + z + 1 = 0$ is $ \langle 1, -1, 1 \rangle $. Therefore, the direction vector of $L_1$ is:
$$ \mathbf{d_1} = \langle 2, -2, 3 \rangle \times \langle 1, -1, 1 \rangle = \langle 1, -1, 0 \rangle $$
Similarly, for the line $L_2$, the normal of the plane $x + 2y - z - 3 = 0$ is $ \langle 1, 2, -1 \rangle $ and for $3x - y + 2z - 1 = 0$ is $ \langle 3, -1, 2 \rangle $. Therefore, the direction vector of $L_2$ is:
$$ \mathbf{d_2} = \langle 1, 2, -1 \rangle \times \langle 3, -1, 2 \rangle = \langle 3, -5, -7 \rangle $$
The normal to the plane containing these two lines can be found as the cross product of $ \mathbf{d_1} $ and $ \mathbf{d_2} $:
$$ \mathbf{n} = \langle 1, -1, 0 \rangle \times \langle 3, -5, -7 \rangle = \langle 7, 7, 2 \rangle $$
The equation of the plane containing the line $L_1$ is obtained by substituting a point on $L_1$. Assume $(3, 4, 7)$ is a point on $L_1$. The equation of the plane is:
$$ 7(x - 3) + 7(y - 4) + 2(z - 7) = 0 $$ which simplifies to $$7x + 7y + 2z = 67$$
Now, calculate the distance from the origin $(0,0,0)$ to this plane using the formula:
$$ D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} $$
Substitute $a = 7, b = 7, c = 2, d = -67$:
$$ D = \frac{|0 + 0 + 0 - 67|}{\sqrt{7^2 + 7^2 + 2^2}} = \frac{67}{\sqrt{49 + 49 + 4}} = \frac{67}{\sqrt{102}} = \frac{67}{3\sqrt{2}} $$
Simplified, the distance $$ D = \frac{1}{3\sqrt{2}} $$.

Therefore, the distance of the origin from the plane containing the lines $L_1$ and $L_2$ is $$ \frac{1}{3\sqrt{2}} $$.

If $L_1$ is the line of intersection of the planes $2x - 2y + 3z - 2 =0, x - y + z + 1 = 0$ and $L_2$ is the line of intersection of the planes $x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0,$ then the distance of the origin from the plane, containing the lines $L_1$ and $L_2$, is:

The Correct Option is B

Solution and Explanation

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