JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

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Application +4 | -1

The line of intersection of the planes
r.

(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=1

and

\vec{r}_{.}(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})=2

, is:

View answer and explanation

Correct option: C

\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{-7}=\frac{z}{-13}

Option C (

\frac{{x-\frac{{6}}{{13}}}}{{2}}=\frac{{y-\frac{{5}}{{13}}}}{{-7}}=\frac{{z}}{{-13}}

) is correct because it represents the line of intersection of the two planes, found by solving the system of equations and finding the direction vector.

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