If the lines $\dfrac{x-1}{-3}=\dfrac{y-2}{2k}=\dfrac{z-3}{2}$ and $\dfrac{x-1}{3k}=\dfrac{y-1}{1}=\dfrac{z-6}{-5}$ are mutually perpendicular, find the value of $k$.

Question

The angle between the lines, whose direction cosines $ l, m, n $ satisfy the equations: \[ l + m + n = 0 \quad \text{and} \quad 2l^2 + 2m^2 - n^2 = 0, \] is:

Answer 1

Step 1: A Vector Based Route:
Write the direction of each line as a vector, then use the fact that perpendicular vectors have zero dot product.
$\vec m=-3\hat i+2k\hat j+2\hat k$ is the direction vector of the first line.
$\vec n=3k\hat i+1\hat j-5\hat k$ is the direction vector of the second line.

Step 2: Set Up the Dot Product Equation:
Perpendicularity means $\vec m\cdot\vec n=0$.
Expand this dot product component by component.
\[ \vec m\cdot\vec n=(-3)(3k)+(2k)(1)+(2)(-5) \]

Step 3: Simplify and Solve:
Multiply out each term.
\[ \vec m\cdot\vec n=-9k+2k-10=-7k-10 \]
Set this equal to zero since the vectors are perpendicular.
\[ -7k-10=0 \]
\[ k=-\dfrac{10}{7} \]

Final Answer:
Both the ratio method and the vector method give the same value of k. \[ \boxed{k=-\dfrac{10}{7}} \]

If the lines \(\dfrac{x-1}{-3}=\dfrac{y-2}{2k}=\dfrac{z-3}{2}\) and \(\dfrac{x-1}{3k}=\dfrac{y-1}{1}=\dfrac{z-6}{-5}\) are mutually perpendicular, find the value of \(k\).

Show Hint

Solution and Explanation

Top Questions on Three Dimensional Geometry

Questions Asked in UP Board XII exam