JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

Let

\hat a{\text{ }}\,\,{\text{and}}\,{\text{ }}\hat b

be two vectors such that

\hat a.\,\,\hat b = \frac{1}{3}

and

\hat a \times \,\hat b = \hat c.

Also

\overline F = \alpha \hat a + \beta \hat b + \gamma \hat c,

where

\alpha ,\,\,\beta ,\,\,\gamma

are scalars. If

\alpha = {k_1}(\hat F.\,\hat a)\, - {k_2}(\hat F.\hat b),

then the value of

2({k_1} + {k_2})

View answer and explanation

Correct option: C

The correct answer is 3 because it is found by expressing F in terms of a, b, and c, then using the properties of the scalar triple product to solve for k1 and k2, and finally calculating 2(k1 + k2).

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}