JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Analyzing +4 | -1

Given four vectors

\mathbf{a}, \mathbf{b}, \mathbf{c}

and

\mathbf{d}

. The vectors

\mathbf{a}, \mathbf{b}, \mathbf{c}

are coplanar but not collinear pair by pair and the vector

\mathbf{d}

is not coplanar with the vecotrs

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

. If it is known that the angle between

\mathbf{a}

and

\mathbf{b}

is equal to that between

b

and c, each being equal to

60^{\circ}

. The angle between

d

and

\mathbf{a}

\alpha

and between

d

and

b

\beta

. Then, the angle between the vectors

d

and

c

View answer and explanation

Correct option: A

\cos ^{-1}(\cos \beta-\cos \alpha)

Option A is correct because the angle between d and c can be found using the given angle conditions and coplanarity of a, b, and c.

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{\hat j}

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\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