JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

1 of 1

Application +4 | -1

Let

\quad \vec{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k} \quad

and

\vec{c}=\hat{i}+2 \hat{j}-2 \hat{k}

where

\alpha, \beta \in \mathbf{R}

, be three vectors. If the projection of

\vec{a}

\vec{c}

\frac{10}{3}

and

\vec{b} \times \vec{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}

, then the value of

\alpha+\beta

is equal to:

View answer and explanation

Correct option: A

Option A (3) is correct because it is found by using the projection formula and the cross product to solve for alpha and beta, then summing them.

More Questions from CROSS PRODUCT

Practice All Chapter MCQs →

\vec{r} \times \vec{b}=\vec{a} \times \vec{b}

\vec{a}, \vec{b}

\lambda

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

x.

\left| {a \times b} \right| = r

a = 2\widehat i + \widehat j - 2\widehat k{\text{ }}\,\,{\text{and }}\,\,b = \widehat i + \widehat j.

a \cdot c = |c|,\,\,|c - a\mid = 2\sqrt 2

a \times b

c

30^\circ ,

|(a \times b) \times c|

\vec{A}=i-2 j-3 k, \vec{B}=2 i+j-k, \vec{C}=i+3 j-2 k

(\vec{A} \times \vec{B}) \times \vec{C}

\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b}

O

S

O A

O C

\frac{\text { area of the quadrilateral } O A B C}{\text { area of } S}