JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Application +4 | -1

\overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+\mu(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})+v(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})

and

[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=\frac{1}{8}

, then

\lambda+\mu+v

is equal to -

View answer and explanation

Correct option: D

8 \overrightarrow{\mathrm{d}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})

Option D (

8 \overrightarrow{{\mathrm{{d}}}} \cdot (\overrightarrow{{\mathrm{{a}}}} + \overrightarrow{{\mathrm{{b}}}} + \overrightarrow{{\mathrm{{c}}}})

) is correct because by taking the dot product of

\overrightarrow{{\mathrm{{d}}}}

with

(\overrightarrow{{\mathrm{{a}}}} + \overrightarrow{{\mathrm{{b}}}} + \overrightarrow{{\mathrm{{c}}}})

and using the given scalar triple product, we can find the value of

\lambda + \mu + \nu

More Questions from DOT PRODUCT OF VECTORS

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O A

O B

30^{\circ}

45^{\circ}

O B

2x + 4y - 5z = 10