If the components of $\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively, are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k}), \quad$ then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is equal to :

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Vector Algebra

If the components of

a = α \hat{i} + β \hat{j} + γ \hat{k}

along and perpendicular to

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

α^{2} + β^{2} + γ^{2}

is equal to :

Answer: (D)

Since

a = a_{∥} + a_{⊥}

a = \frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k}) + \frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k}) = \frac{1}{11} [(48 - 4) \hat{i} + (16 - 5) \hat{j} + (- 16 - 17) \hat{k}]

= \frac{1}{11} (44 \hat{i} + 11 \hat{j} - 33 \hat{k}) = 4 \hat{i} + \hat{j} - 3 \hat{k}

α = 4, β = 1, γ = - 3

, and

α^{2} + β^{2} + γ^{2} = 16 + 1 + 9 = 26

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option