JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Analyzing +4 | -1

The vector

\vec{a}=\hat{i}+2 \hat{j}+\hat{k}

lying in the plane of the vectors

\vec{b}=\hat{i}+\hat{j}

and

\vec{c}

bisects the angle between

\vec{b}

and

\vec{c}

. If

|\vec{c}|=\sqrt{2}

, then

\vec{c} \cdot \vec{a}

View answer and explanation

Correct option: C

Option C (3) is correct because by expressing

\vec{a}

as a linear combination of

\frac{\vec{b}}{|\vec{b}|}

and

\vec{c}

, and using the given conditions, we find

\vec{c} = \hat{j} + \hat{k}

, so

\vec{c} \cdot \vec{a} = 3

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