JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

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Analyzing +4 | -1

Let

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

be three non-coplanar vectors and

\mathbf{d}

\mathbf{a}

non-zero vector, which is perpendicular to

\mathbf{a}+\mathbf{b}+\mathbf{c}

Now, if

\mathbf{d}=(\sin x)(\mathbf{a} \times \mathbf{b})+(\cos y)(\mathbf{b} \times \mathbf{c})+2(\mathbf{c} \times \mathbf{a})

, then minimum value of

x^2+y^2

is equal to

View answer and explanation

Correct option: D

\frac{5 \pi^2}{4}

Option D (

\frac{{5 \pi^2}}{4}

) is correct because it is the minimum value of x² + y² that satisfies the given conditions.

More Questions from 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}