let a11= component of a along b a1= component of a perpendicular to b $\begin{aligned}
& \overrightarrow{\mathrm{a}}_{11}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \
& \overrightarrow{\mathrm{a}}1=\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \
& \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}{11}+\overrightarrow{\mathrm{a}}_1
\end{aligned}\begin{aligned} & \begin{array}{l}\therefore \overrightarrow{\mathrm{a}}= \ \quad=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \ \overrightarrow{11} \hat{\mathrm{j}}-\frac{33}{11} \hat{\mathrm{k}}\end{array} \ & \begin{aligned} \alpha=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}}\end{aligned} \ & \alpha^2+\beta^2+\gamma^2=16+1+9=26\end{aligned}$
If the components of overrightarrow a=α i+β j+γ k along and perpendicular to overrightarrow b=3 i+ j- k respectively, are 16 11(3 i+ j- k) and 1…
JEE Main 2025 — Mathematics Vectors & 3D Geometry
If the components of a=αi^+βj^+γk^ along and perpendicular to b=3i^+j^−k^ respectively, are 1116(3i^+j^−k^) and 111(−4i^−5j^−17k^), then α2+β2+γ2 is equal to :
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
26
- B
18
- C
23
- D
16
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