JEE Mains · Maths · STD 12 - 10. vector algebra
A vector \(\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}(\alpha, \beta \in \mathrm{R})\) lies in the plane of the vectros \(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}\) and \(\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) If \(\overrightarrow{\mathrm{a}}\) bisects the angle between \(\overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{c}},\) then
- A \(\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{i}}+1=0\)
- B \(\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{i}}+3=0\)
- C \(\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{k}}+4=0\)
- D \(\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{k}}-4=0\)
Answer & Solution
Correct Answer
(D) \(\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{k}}-4=0\)
Step-by-step Solution
Detailed explanation
\(\vec{a}=\lambda(\hat{b}+\hat{c})=\lambda\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}+\frac{\hat{i}-\hat{j}+4 \hat{k}}{3 \sqrt{2}}\right)\) \(\vec{a}=\frac{\lambda}{3 \sqrt{2}}(4 \hat{i}+2 \hat{j}+4 \hat{k}) \Rightarrow \frac{\lambda}{3 \sqrt{2}}(4 \hat{i}+2 \hat{j}+4 \hat{k})\)…
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