If the components of vector a=αi+βj+γk along and perpendicular to vector b=3i+j-k respectively,

Question

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

(1) 23

(2) 18

(3) 16

(4) 26

Answer 1

Correct option is (4) 26  

Let

\(\overrightarrow{a}_{11}=\) component of \(\overrightarrow{a}\) along \(\overrightarrow{b}\)

\(\overrightarrow{a}_{1}=\) component of \(\vec{a}\) perpendicular to \(\overrightarrow{b}\)

\(\overrightarrow{\mathrm{a}}_{11}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})\)

\(\overrightarrow{\mathrm{a}}_{1}=\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}})\)

\(\because \overrightarrow{a}=\overrightarrow{a}_{11}+\overrightarrow{a}_{1}\)

\(\therefore \overrightarrow{\mathrm{a}}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}})\)

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

\(\overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)

\(\alpha=4 \quad \beta=1 \quad \gamma=-3\)

\(\alpha^{2}+\beta^{2}+\gamma^{2}=16+1+9=26\)