JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Let two non-collinear vectors

\mathbf{a}

and

\mathbf{b}

inclined at an angle

\frac{2 \pi}{3}

be such that

|\mathbf{a}|=3

and

|\mathbf{b}|=2

. If a point

P

moves so that at any time

t

its position vector

\mathrm{OP}

(where

O

is the origin) is given as

\mathbf{O P}=\left(t+\frac{1}{t}\right) \mathbf{a}+\left(t-\frac{1}{t}\right) \mathbf{b}

, then least distance of

P

from the origin is

View answer and explanation

Correct option: B

\sqrt{2 \sqrt{133}+10}

Option B (

\sqrt{2 \sqrt{133}+10}

) is correct because it represents the minimum distance of point P from the origin, obtained by minimizing the square of the distance using AM-GM inequality.

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