JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

Let

{x_0}

be the point of local maximum of

f(x) = a.(b \times c),

where

a = x\widehat i - 2\widehat j + 3\widehat k,b = - 2\widehat i + x\widehat j - \widehat k\mathrm{ }\,\mathrm{and }\,\,c = 7\widehat i - 2\widehat j + x\widehat k

Then the value

a.\,b + b.\,c + c.\,a

x = {x_0}

View answer and explanation

Correct option: C

-22

Option C (-22) is correct because calculating the scalar triple product, finding the local maximum, and substituting it into the expression yields -22.

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}