KCET · Maths · Vector Algebra
The value of \(\lambda\) for which the vectors \(\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\) are orthogonal is
- A \(\dfrac{5}{2}\)
- B \(\dfrac{-5}{2}\)
- C \(\dfrac{2}{5}\)
- D \(\dfrac{-2}{5}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{-5}{2}\)
Step-by-step Solution
Detailed explanation
For two vectors \(\vec{a}\) and \(\vec{b}\) to be orthogonal, their dot product must be zero.
\(\vec{a} \cdot \vec{b} = 0\)
\((2\hat{i} + \lambda\hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 0\)
\((2)(1) + (\lambda)(2) + (1)(3) = 0\)
\(2 + 2\lambda + 3 = 0\)
\(2\lambda + 5 = 0\)
\(\lambda = \dfrac{-5}{2}\)
Answer: \(\dfrac{-5}{2}\)
\(\vec{a} \cdot \vec{b} = 0\)
\((2\hat{i} + \lambda\hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 0\)
\((2)(1) + (\lambda)(2) + (1)(3) = 0\)
\(2 + 2\lambda + 3 = 0\)
\(2\lambda + 5 = 0\)
\(\lambda = \dfrac{-5}{2}\)
Answer: \(\dfrac{-5}{2}\)
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