JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

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Application +4 | -1

The sine of the angle between the two vectors

3 \mathbf{i}+2 \mathbf{j}-\mathbf{k}

and

12 \mathbf{i}+5 \mathbf{j}-5 \mathbf{k}

will be

View answer and explanation

Correct option: A

\frac{\sqrt{115}}{\sqrt{14} \sqrt{194}}

Option A (

\frac{{\sqrt{{115}}}}{{\sqrt{{14}} \sqrt{{194}}}}

) is correct because it represents the sine of the angle between the two vectors, calculated using the magnitude of their cross product divided by the product of their magnitudes.

More Questions from CROSS PRODUCT

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A solution of the vector equation

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

, where

\vec{a}, \vec{b}

are two given vectors, is where

\lambda

is a parameter. Let

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

for some real

x.

Then

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

is possible if A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(-1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is Let

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

If c is a vector such that

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

and the angle between

a \times b

and

c

30^\circ ,

then

|(a \times b) \times c|

equal to If

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

, then

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

is Let

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

, where

O

is the origin, If

S

is the parallelogram with adjacent sides

O A

and

O C

, then

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

is equal to