JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

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Analyzing +4 | -1

Let

\vec{a}

be a non-zero vector parallel to the line of intersection of the two planes described by

\hat{i}+\hat{j}, \hat{i}+\hat{k}

and

\hat{i}-\hat{j}, \hat{j}-\hat{k}

. If

\theta

is the angle between the vector

\vec{a}

and the vector

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

and

\vec{a} \cdot \vec{b}=6

, then the ordered pair

(\theta,|\vec{a} \times \vec{b}|)

is equal to

View answer and explanation

Correct option: D

\left(\frac{\pi}{4}, 6\right)

Option D is correct. By finding the direction vector of the line of intersection, then using the dot product formula, we can find the angle and the magnitude of the cross product.

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