JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

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Application +4 | -1

Area of rectangle having vertices A, B, C and D with position vector

\left(-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\right),\left(\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\right),\left(\hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\right)

and

\left(-\hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\right)

View answer and explanation

Correct option: A

2 sq unit

Option A (2 sq unit) is correct because the area of the rectangle is calculated by finding the lengths of the sides using the position vectors and then multiplying them.

More Questions from DOT PRODUCT OF VECTORS

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The distance between the planes given by

\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0

and

\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0

is A vector which makes equal angles with the vectors

\frac{1}{3}\left( {\hat i - 2\hat j + 2\hat k} \right),\frac{1}{5}\left( { - 4\hat i - 3\hat k} \right)

, and

{\hat j}

, is If vector

\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

and vector

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

, then

\frac{\text { Projection of vector } \mathbf{a} \text { on vector } \mathbf{b}}{\text { Projection of vector } \mathbf{b} \text { on vector } \mathbf{a}}

is equal to The angle between

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

\frac{{5\pi }}{6}

and the projection of

{\mathbf{\vec a}}

in the direction of

{\mathbf{\vec b}}

\frac{{ - 6}}{{\sqrt 3 }}

then

\left| {\mathbf{\vec a}} \right|

is equal to A velocity

\frac{1}{4} \mathrm{~m} / \mathrm{s}

is resolved into two components along

O A

and

O B

making angles

30^{\circ}

and

45^{\circ}

respectively with the given velocity. Then the component along

O B

is Find the equation of the perpendicular drown from the origin to the plane

2x + 4y - 5z = 10