KCET · Maths · Vector Algebra
If \(\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}, \vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}\) and \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\), then \(\alpha + \beta\) is equal to
- A \(2\)
- B \(-1\)
- C \(0\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
Given \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\)
Squaring both sides:
\(|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}\)
\(4\vec{a} \cdot \vec{b} = 0\)
\(\vec{a} \cdot \vec{b} = 0\)
Substituting the given vectors \(\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}\) and \(\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}\):
\((2)(\alpha) + (2)(\beta) + (-1)(2) = 0\)
\(2\alpha + 2\beta - 2 = 0\)
\(2(\alpha + \beta) = 2\)
\(\alpha + \beta = 1\)
Answer: \(1\)
Squaring both sides:
\(|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}\)
\(4\vec{a} \cdot \vec{b} = 0\)
\(\vec{a} \cdot \vec{b} = 0\)
Substituting the given vectors \(\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}\) and \(\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}\):
\((2)(\alpha) + (2)(\beta) + (-1)(2) = 0\)
\(2\alpha + 2\beta - 2 = 0\)
\(2(\alpha + \beta) = 2\)
\(\alpha + \beta = 1\)
Answer: \(1\)
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