Question

Let \[\vec{a} = a_1 \hat { i }+ a_2 \hat {j} + a_3 \hat {k} , \vec{b} = b_1 \hat {i} + b_2 \hat { j } + b_3 \hat { k} \text { and } \vec{c} = c_1 \hat { i } + c_2 \hat{j } + c_3\text {  k }\] be three non-zero vectors such that \[\vec{c}\] is a unit vector perpendicular to both \[\vec{a} \text { and } \vec{b}\]. If the angle between \[\vec{a} \text { and } \vec{b}\] is \[\frac{\pi}{6},\] , then

\[\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}^2\] is equal to

विकल्प

• 0

• 1

• \[\frac{1}{4} \left| \vec{a} \right|^2 \left| \vec{b} \right|^2\]

• \[\frac{3}{4} \left| \vec{a} \right|^2 \left| \vec{b} \right|^2\]

Answer 1

\[\frac{1}{4} \left| \vec{a} \right|^2 \left| \vec{b} \right|^2\]

We have

\[ \begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}^2 \]

\[ = \left[ \left( \vec{a} \times \vec{b} \right) . \vec{c} \right]^2 \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \left( \text { By definition of scalar triple product } \right)\]

\[ = \left[ \left| \left( \vec{a} \times \vec{b} \right) \right|\left| \vec{c} \right|cos0^\circ \right]^2 \left( \because \vec{a} \times \vec{b} \text { is parallel to vector } \vec{c} \text { as } \vec{c} \text { is perpendicular to both } \vec{ a } \text { and } \vec{ b } \right)\]

\[ = \left( \left| \vec{a} \right|\left| \vec{b} \right|sin\frac{\pi}{6} \right)^2 \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \left( \because \left| \vec{c} \right| = 1 \text { and angle between } \vec{a} \text { and } \vec{b} is \frac{\pi}{6} \right)\]

\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \left( \frac{1}{2} \right)^2 \]

\[ = \frac{1}{4} \left| \vec{a} \right|^2 \left| \vec{b} \right|^2\]