Question

The point of intersection \(C\) of the plane \(8 x+y+2 z=0\) and the line joining the points \(A (-3,-6,1)\) and \(B (2,4,-3)\)divides the line segment \(AB\) internally in the ratio\(k : 1 \  If a , b , c (| a |,| b |, | c |\)are coprime) are the direction ratios of the perpendicular from the point \(C\)on the line \(\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}\), then \(| a + b + c |\)is equal to ___

Hint:

To find the perpendicular from a point to a line, calculate the direction vector from the point to a general point on the line and ensure it is perpendicular to the line’s direction vector using the dot product.

Answer 1

Correct Answer: 10

Step 1: Equation of Line \(AB\)
The line passing through \(A(-3, -6, 1)\) and \(B(2, 4, -3)\) is given by:
\[\frac{x - 2}{5} = \frac{y - 4}{10} = \frac{z + 3}{-4} = \lambda.\]
Step 2: Intersection of Line and Plane
Any point on the line is:
\[(5\lambda + 2, 10\lambda + 4, -4\lambda - 3).\]
Substitute this into the plane equation \(8x + y + 2z = 0\):
\[8(5\lambda + 2) + (10\lambda + 4) + 2(-4\lambda - 3) = 0.\]
\[40\lambda + 16 + 10\lambda + 4 - 8\lambda - 6 = 0.\]
\[42\lambda + 14 = 0 \implies \lambda = -\frac{1}{3}.\]
The point \(C\) is:
\[C = \left(\frac{1}{3}, \frac{2}{3}, -\frac{5}{3}\right).\]
Step 3: Perpendicular from \(C\) to Line \(L\)
The line \(L\) is given by:
\[\frac{x - 1}{-1} = \frac{y + 4}{2} = \frac{z + 2}{3} = \mu.\]
A general point on \(L\) is:
\[(-\mu + 1, 2\mu - 4, 3\mu - 2).\]
Direction vector of \(\overrightarrow{CD}\) is:
\[\overrightarrow{CD} = \left(-\mu + 1 - \frac{1}{3}\right)\mathbf{i} + \left(2\mu - 4 - \frac{2}{3}\right)\mathbf{j} + \left(3\mu - 2 + \frac{5}{3}\right)\mathbf{k}.\]
\[\overrightarrow{CD} = \left(-\mu + \frac{2}{3}\right)\mathbf{i} + \left(2\mu - \frac{14}{3}\right)\mathbf{j} + \left(3\mu - \frac{1}{3}\right)\mathbf{k}.\]
Since \(\overrightarrow{CD}\) is perpendicular to the line \(L\), the dot product of \(\overrightarrow{CD}\) with \((-1, 2, 3)\) is zero:
\[\left(-\mu + \frac{2}{3}\right)(-1) + \left(2\mu - \frac{14}{3}\right)(2) + \left(3\mu - \frac{1}{3}\right)(3) = 0.\]
\[\mu - \frac{2}{3} + 4\mu - \frac{28}{3} + 9\mu - \frac{1}{3} = 0.\]
\[14\mu - \frac{31}{3} = 0 \implies \mu = \frac{11}{14}.\]
Substituting \(\mu = \frac{11}{14}\) into \(\overrightarrow{CD}\), the direction ratios of \(\overrightarrow{CD}\) are:
\[(-1, -26, 17).\]
Step 4: Calculate \(|a + b + c|\)
\[|a + b + c| = |-1 - 26 + 17| = 10.\]
Conclusion: The value of \(|a + b + c|\) is \(10\).