Let a =a_1 i +a_2 j +a_3 k , b =b_1 i +b_2 j +b_3 k and c =c_1 i +c_2 j +c_3 k be three zero vectors such that c is a unit vector perpendicular to both a and b . If the angle between a and b is 6 , then a_1&a_2&a_3 _1&b_2&b_3 _1&c_2&c_3 ^2 is equal to: 1. 0 2. 1 3. 1 4 | a |^2 | b |^2 4. 3 4 | a |^2 | b |^2

3. 1 4 | a |^2 | b |^2 Solution: We have a_1&a_2&a_3 _1&b_2&b_3 _1&c_2&c_3 ^2 = [ ( a b ). c ]^2 (By definition of scalar triple product) = [ | ( a b ) | | c | 0^ ]^2 ( a b is parallel to vector c as c is perpendicular to both a and b ) = ( | a | | b | 6 )^2 ( | c |=1 and angle between a and b is 6 ) = | a |^2 | b |^2 (1 2 )^2 =1 4 | a |^2 | b |^2

Let a =a_1 i +a_2 j +a_3 k , b =b_1 i +b_2 j +b_3 k and c =c_1 i …

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Question

Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ be three zero vectors such that $\vec{\text{c}}$ is a unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$ If the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6},$ then $\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}^2$ is equal to:

$0$
$1$
$\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
$\frac{3}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

✓

Answer

$\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

Solution:

We have

$\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}^2$

$=\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big]^2$ (By definition of scalar triple product)

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

$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\frac{\pi}{6}\big)^2$ $\big(\therefore\big|\vec{\text{c}}\big|=1$ and angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6}\big)$

$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\frac{1}{2}\big)^2$

$=\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

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