JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}

, then the value of

\left|\begin{array}{ccc}\mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c}\end{array}\right|

View answer and explanation

Correct option: C

Option C (16) is correct because the determinant can be simplified to the square of the scalar triple product of the vectors a, b, and c, which evaluates to 16.

More Questions from SCALAR TRIPLE PRODUCT

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

for some non-zero vector

\vec r,

then the value of

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

is The shortest distance between the lines

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

and

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

is Let

\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}

and

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

Then

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

depends on If

\vec{a}, \vec{b}

are non-zero vectors, then

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

equals

a \cdot(a \times b)=

The length of the perpendicular from the origin to the plane passing through the point a and containing the line

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