If $a = \widehat i + \widehat j + \widehat k,\,\,\,\,b = \widehat i - \widehat j + \widehat k,\,\,\,c = \widehat i + 2\widehat j - \widehat k,$ then the value of $\left| {\begin{array}{*{20}{c}} {a \cdot a}&{a \cdot b}&{a \cdot c} \\ {b \cdot a}&{b \cdot b}&{b \cdot c} \\ {c \cdot a}&{c \cdot b}&{c \cdot c} \end{array}} \right|\,$ is

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

a = \widehat i + \widehat j + \widehat k,\,\,\,\,b = \widehat i - \widehat j + \widehat k,\,\,\,c = \widehat i + 2\widehat j - \widehat k,

then the value of

\left| {\begin{array}{cccccccccccccccccccc} {a \cdot a}&{a \cdot b}&{a \cdot c} \\ {b \cdot a}&{b \cdot b}&{b \cdot c} \\ {c \cdot a}&{c \cdot b}&{c \cdot c} \end{array}} \right|\,

View answer and explanation

Correct option: C

Option C (16) is correct because the determinant of the matrix formed by dot products is equal to the square of the scalar triple product.

Practice All Chapter MCQs →

\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

\vec r,

\left[ {\vec a\vec b\vec c} \right],

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

\mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k}

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

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

\vec{a}, \vec{b}

((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b})

a \cdot(a \times b)=

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}