JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Analyzing +4 | -1

Let

\hat{a}

and

\hat{b}

be two unit vectors such that the angle between them is

\frac{\pi}{4}

. If

\theta

is the angle between the vectors

(\hat{a}+\hat{b})

and

(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))

, then the value of

164 \cos ^2 \theta

is equal to :

View answer and explanation

Correct option: A

90+27 \sqrt{2}

Option A (

90+27 \sqrt{2}

) is correct because it is the value of

164 \cos^2 \theta

calculated using the dot product and magnitudes of the given vectors.

More Questions from DOT PRODUCT OF VECTORS

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

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

\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