ExamBro
ExamBro
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 :

  1. A \(26\)
  2. B \(18\)
  3. C \(23\)
  4. D \(16\)
Verified Solution

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 \(\vec{b}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app