Question

The value of \(λ\) for which the vectors \(i+j-k\)and \(λi+3j+k\) are perpendicular is ?

• A. \(-2\)
• B. \(2\)
• C. \(0\)
• D. \(-1\)
• E. \(1\)

Answer 1

The Correct Option is A

Step 1: Recall that two vectors are perpendicular if and only if their dot product equals zero.
Given vectors:
\[ \vec{a} = \hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = \lambda\hat{i} + 3\hat{j} + \hat{k} \]

Step 2: Compute the dot product \(\vec{a} \cdot \vec{b}\):
\[ \vec{a} \cdot \vec{b} = (1)(\lambda) + (1)(3) + (-1)(1) = \lambda + 3 - 1 = \lambda + 2 \]

Step 3: Set the dot product equal to zero for perpendicular vectors and solve for \(\lambda\):
\[ \lambda + 2 = 0 \]
\[ \lambda = -2 \]

Conclusion: The value of \(\lambda\) that makes the vectors perpendicular is \(\boxed{-2}\).

Answer 2

Two vectors are perpendicular if their dot product is zero. Let's find the dot product of the given vectors:

\[ (\mathbf{i} + \mathbf{j} - \mathbf{k}) \cdot (\lambda\mathbf{i} + 3\mathbf{j} + \mathbf{k}) = 0 \]

The dot product is calculated by multiplying corresponding components and summing the results:

\[ 1 \cdot \lambda + 1 \cdot 3 + (-1) \cdot 1 = 0 \] \[ \lambda + 3 - 1 = 0 \] \[ \lambda + 2 = 0 \] \[ \lambda = -2 \]

Therefore, the value of \( \lambda \) for which the vectors are perpendicular is \(-2\).