Question:
The set of real values of \(\lambda\) for which the vectors
\[
\lambda \hat{i}-3\hat{j}+5\hat{k}
\]
and
\[
2\lambda \hat{i}-\lambda \hat{j}+\hat{k}
\]
are perpendicular to each other is
The set of real values of \(\lambda\) for which the vectors
\[
\lambda \hat{i}-3\hat{j}+5\hat{k}
\]
and
\[
2\lambda \hat{i}-\lambda \hat{j}+\hat{k}
\]
are perpendicular to each other is
Show Hint
Two vectors are perpendicular if their dot product is zero. After forming the resulting equation, check the discriminant to determine whether real solutions exist.
Updated On: Jun 26, 2026
- \(\{0,1\}\)
- \(\{-2\}\)
- \(\{2,-1\}\)
- \(\phi\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use the condition for perpendicular vectors.
Two vectors are perpendicular if and only if their dot product is zero.
Let \[ \vec{a}=(\lambda,-3,5) \] and \[ \vec{b}=(2\lambda,-\lambda,1). \] Therefore, \[ \vec{a}\cdot \vec{b}=0. \]
Step 2: Compute the dot product.
\[ (\lambda)(2\lambda)+(-3)(-\lambda)+(5)(1)=0 \] \[ 2\lambda^2+3\lambda+5=0. \]
Step 3: Check for real solutions.
For the quadratic equation \[ 2\lambda^2+3\lambda+5=0, \] the discriminant is \[ D=b^2-4ac. \] Thus, \[ D=3^2-4(2)(5) \] \[ D=9-40 \] \[ D=-31. \]
Step 4: Analyze the discriminant.
Since \[ D\lt 0, \] the quadratic equation has no real roots.
Hence, there is no real value of \(\lambda\) for which the given vectors are perpendicular.
Step 5: Final conclusion.
Therefore, \[ \boxed{\phi} \] and the correct option is \[ \boxed{(4)}. \]
Two vectors are perpendicular if and only if their dot product is zero.
Let \[ \vec{a}=(\lambda,-3,5) \] and \[ \vec{b}=(2\lambda,-\lambda,1). \] Therefore, \[ \vec{a}\cdot \vec{b}=0. \]
Step 2: Compute the dot product.
\[ (\lambda)(2\lambda)+(-3)(-\lambda)+(5)(1)=0 \] \[ 2\lambda^2+3\lambda+5=0. \]
Step 3: Check for real solutions.
For the quadratic equation \[ 2\lambda^2+3\lambda+5=0, \] the discriminant is \[ D=b^2-4ac. \] Thus, \[ D=3^2-4(2)(5) \] \[ D=9-40 \] \[ D=-31. \]
Step 4: Analyze the discriminant.
Since \[ D\lt 0, \] the quadratic equation has no real roots.
Hence, there is no real value of \(\lambda\) for which the given vectors are perpendicular.
Step 5: Final conclusion.
Therefore, \[ \boxed{\phi} \] and the correct option is \[ \boxed{(4)}. \]
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