JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Application +4 | -1

If two adjacent sides of two rectangles are represented by the vectors

\mathbf{p}=5 \mathbf{a}-3 \mathbf{b}, \mathbf{q}=-\mathbf{a}-2 \mathbf{b}

and

r=-4 a-b ; s=-a+b

respectively, then the angle between the vectors

\mathbf{x}=\frac{1}{3}(\mathrm{p}+\mathbf{r}+s)

and

\mathrm{y}=\frac{1}{5}(\mathbf{r}+s)

View answer and explanation

Correct option: B

\cos ^{-1}\left(\frac{19}{5 \sqrt{43}}\right)

Option B (

\cos ^{-1}\left(\frac{19}{5 \sqrt{43}}\right)

) is correct because it is the angle calculated using the dot product of the simplified vectors x and y.

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