JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Analyzing +4 | -1

a

is a real constant and

\mathrm{A}, \mathrm{B}, \mathrm{C}

are variable angles and

\sqrt{a^2-4} \operatorname{Tan} A+a \operatorname{Tan} B+\sqrt{a^2+4} \operatorname{Tan} C=6 a

Then the least value of

\operatorname{Tan}^2 A+\operatorname{Tan}^2 B+\operatorname{Tan}^2 C=

View answer and explanation

Correct option: C

Option A (10,11) is correct because the number of solutions depends on the range of the greatest integer function, and the given conditions are satisfied when n is 10 or 11.

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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