Question

The length of the perpendicular from the point $(1,-2,5)$ on the line passing through $(1,2,4)$ and parallel to the line $x+y-z=0=x-2 y+3 z-5$ is :

• A. $\sqrt{\frac{21}{2}}$
• B. $\sqrt{\frac{9}{2}}$
• C. $\sqrt{\frac{73}{2}}$
• D. 1

Answer 1

The Correct Option is A

To find the length of the perpendicular from the point \((1, -2, 5)\) onto the line passing through \((1, 2, 4)\) and is parallel to the given plane system, we need to follow these steps:

1. First, determine the direction vector of the line. The line is parallel to the intersection of two planes \(x + y - z = 0\) and \(x - 2y + 3z - 5 = 0\). The direction of the line of intersection of two planes can be found by taking the cross product of the normal vectors of the planes.
2. Find the normals: The normal vector of the plane \(x + y - z = 0\) is \((1, 1, -1)\), and the normal vector of the plane \(x - 2y + 3z - 5 = 0\) is \((1, -2, 3)\).
3. Compute the cross product of these normal vectors to get the direction vector of the line:
   \[ \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 1 & -2 & 3 \end{vmatrix} = \mathbf{i}(1 \cdot 3 - (-1) \cdot (-2)) - \mathbf{j}(1 \cdot 3 - (-1) \cdot 1) + \mathbf{k}(1 \cdot (-2) - 1 \cdot 1) \] \[ = \mathbf{i}(3 - 2) - \mathbf{j}(3 + 1) + \mathbf{k}(-2 - 1) = \mathbf{i}(1) - \mathbf{j}(4) + \mathbf{k}(-3) \] \]
   Thus, the direction vector \(\mathbf{v} = (1, -4, -3)\).
4. Now, using the given point on the line \((1, 2, 4)\) and the direction vector \(\mathbf{v} = (1, -4, -3)\), the parametric equation of the line can be expressed as:
   \[ (x, y, z) = (1, 2, 4) + t(1, -4, -3) \] \
5. The vector from the point \((1, -2, 5)\) to the point on the line \((1, 2, 4)\) is \((0, 4, -1)\).
6. The perpendicular distance \(d\) from a point to a line in space is given by the formula:
   \[ d = \frac{\|\mathbf{AP} \times \mathbf{v}\|}{\|\mathbf{v}\|} \]
   where \(\mathbf{AP}\) is the vector from the point \((1, 2, 4)\) to the point \((1, -2, 5)\) (i.e., \((0, 4, -1)\)).
7. Find the cross product \(\mathbf{AP} \times \mathbf{v}\):
   \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 4 & -1 \\ 1 & -4 & -3 \end{vmatrix} = \mathbf{i}(-12 + 4) - \mathbf{j}(-3 - 1) + \mathbf{k}(0 - 4(1)) \] \[ = \mathbf{i}(-8) + \mathbf{j}(4) - \mathbf{k}(4) = (-8, 4, -4) \] \]
   Thus, the result of the cross product is \((-8, 4, -4)\).
8. Calculate the magnitude of the cross product and direction vector:
   \[ \|\mathbf{AP} \times \mathbf{v}\| = \sqrt{(-8)^2 + 4^2 + (-4)^2} = \sqrt{64 + 16 + 16} = \sqrt{96} \] \[ \|\mathbf{v}\| = \sqrt{1^2 + (-4)^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \]
9. The length of the perpendicular from the point can be calculated as:
   \[ d = \frac{\sqrt{96}}{\sqrt{26}} = \frac{\sqrt{96}}{\sqrt{26}} = \sqrt{\frac{96}{26}} = \sqrt{\frac{48}{13}} = \sqrt{\frac{21}{2}} \]

Therefore, the length of the perpendicular from the point \((1, -2, 5)\) to the line is \(\sqrt{\frac{21}{2}}\).