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The vector $\vec{a}=\alpha \, \vec{i}+\, 2\vec{j}+\, \beta \vec{k}$ lies in the plane of the vectors $\vec{b}= \vec{i}+\, \vec{j}\; \; and\; \; \vec{c}= \vec{j}+\, \vec{k}\; \;$ and bisects the angle between $\vec{b}\; \; and\; \; \vec{c}$ . Then which one of the following gives possible values of $\alpha \; \; and\; \; \beta$ ?

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

Option 2)
$\alpha =2,\beta =2$

Option 3)
$\alpha =1,\beta =2(d)$

Option 4)
$\alpha =2,\beta =1$

Answers (1)

best_answer

As we have learned

Scalar Triple Product -

$\left [ \vec{a}\;\vec{b}\; \vec{c} \right ]$

$=\left (\vec{a}\times \vec{b}\right)\cdot \vec{c}= \vec{a}\cdot \left ( \vec{b} \times \vec{c}\right )$

$=\left (\vec{b}\times \vec{c}\right)\cdot \vec{a}= \vec{b}\cdot \left ( \vec{c} \times \vec{a}\right )$

$=\left (\vec{c}\times \vec{a}\right)\cdot \vec{b}= \vec{c}\cdot \left ( \vec{a} \times \vec{b}\right )$

- wherein

Scalar Triple Product of three vectors $\hat{a},\hat{b},\hat{c}$ .

$\vec{a}\cdot (\vec{b}\times \vec{c}) = 0$

$\Rightarrow \begin{vmatrix} \alpha & 2 &\beta \\ 1& 1& 0\\ 0&1 &1 \end{vmatrix}= 0$

$\alpha -2+\beta = 0 ............(1 )$

$\alpha\hat i + 2\hat j +\beta \hat k = \lambda\left ( \frac{\vec{b}}{|\vec{b}|}+\frac{\vec{c}}{|\vec{c}|} \right )$

$= \lambda \left ( \frac{\hat{i}+2\hat{j}+\hat{k}}{\sqrt 2 } \right )$

$\alpha = \lambda /\sqrt 2 \\ 2 = \sqrt 2 \lambda \\ \beta = \lambda /sqrt 2 \\ \alpha =\beta =1$

Option 1)

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

Option 2)

$\alpha =2,\beta =2$

Option 3)

$\alpha =1,\beta =2(d)$

Option 4)

$\alpha =2,\beta =1$

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Himanshu

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