JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Analyzing +4 | -1

Let

\overrightarrow a

and

\overrightarrow b

be the vectors along the diagonals of a parallelogram having area

2\sqrt 2 .

Let the angle between

\overrightarrow a

and

\overrightarrow b

be acute,

\left| {\overrightarrow a } \right| = 1,

and

\left| {\overrightarrow a \cdot \overrightarrow b } \right| = \left| {\overrightarrow a \times \overrightarrow b } \right|.

\overrightarrow c = 2\sqrt 2 \left( {\overrightarrow a \times \overrightarrow b } \right) - 2\overrightarrow b ,

then an angle between

\overrightarrow b

and

\overrightarrow c

View answer and explanation

Correct option: D

\frac{{3\pi }}{4}

Option D (

\frac{{3\pi }}{{4}}

) is correct because it is the angle between vectors

\overrightarrow b

and

\overrightarrow c

calculated using the given conditions and vector properties.

More Questions from DOT PRODUCT OF VECTORS

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

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

{\hat j}

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

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

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

{\mathbf{\vec a}}

{\mathbf{\vec b}}

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

{\mathbf{\vec a}}

{\mathbf{\vec b}}

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

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

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

O A

O B

30^{\circ}

45^{\circ}

O B

2x + 4y - 5z = 10