MCQ
$\vec a $ એ $\vec b$ તથા $\vec c$ વચ્ચેના ખૂણાનો દુભાજક હોય તથા $\vec a = \left( {\alpha ,2,\beta } \right)\;,\vec b = \left( {1,1,0} \right),\;\vec c = \left( {0,1,1} \right)$ તો $\alpha ,\beta $ ની શક્ય કિંમતો મેળવો.
  • A
    $\alpha = 2,\beta = 2$
  • B
    $\;\alpha = - 1,\beta = 1$
  • C
    $\;\alpha = 2,\beta = 1$
  • $\;\alpha = 1,\beta = 1$

Answer

Correct option: D.
$\;\alpha = 1,\beta = 1$
d
Given: $\overrightarrow{\boldsymbol{a}}=\alpha \hat{\boldsymbol{i}}+2 \hat{\boldsymbol{j}}+\beta \hat{\boldsymbol{k}}, \overrightarrow{\boldsymbol{b}}=\hat{\boldsymbol{i}}+\hat{\boldsymbol{j}} \overrightarrow{\boldsymbol{c}}=\hat{\boldsymbol{j}}+\hat{\boldsymbol{k}}$ are coplanar.

$\Rightarrow[\vec{a} \vec{b} \vec{c}]=0$

$\Rightarrow\left|\begin{array}{lll}{\alpha} & {2} & {\beta} \\ {1} & {1} & {0} \\ {0} & {1} & {1}\end{array}\right|=0$

$\Rightarrow \alpha-2+\beta=0$ or $\alpha+\beta=2 \dots$    . .$(i)$

Also it is given that $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$ $\Rightarrow \frac{\alpha+2}{|\vec{a}||\vec{b}|}=\frac{2+\beta}{|\vec{a}||\vec{c}|}$

$\Rightarrow \alpha=\beta $  .....$(ii)$

From $(i)$ and $(ii)$ 

$(\alpha, \beta)=(1,1)$

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