JEE Mains · Maths · STD 12 - 10. vector algebra
If the components of \(\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\) along and perpendicular to \(\overrightarrow{\mathrm{b}}=3 \hat{i}+\hat{j}-\hat{k}\) respectively, are \(\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})\) and \(\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})\), then \(\alpha^2+\beta^2+\gamma^2\) is equal to :
- A \(26\)
- B \(18\)
- C \(23\)
- D \(16\)
Answer & Solution
Correct Answer
(A) \(26\)
Step-by-step Solution
Detailed explanation
let \(\vec{a}_{11}=\text { component of } \vec{a} \text { along } \vec{b}\) \(\vec{a}_1=\) component of \(\vec{a}\) perpendicular to \(\vec{b}\)…
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