JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

1 of 1

Application +4 | -1

A vector

\vec{v}

in the first octant is inclined to the

x

-axis at

60^{\circ}

, to the

y

-axis at

45^{\circ}

and to the

z

-axis at an acute angle. If a plane passing through the points

(\sqrt{2},-1,1)

and

(a, b, c)

is normal to

\vec{v}

, then

View answer and explanation

Correct option: B

a+\sqrt{2 b}+c=1

Option B (

a+\sqrt{{2 b}}+c=1

) is correct because the direction cosines of the vector

\vec{{v}}

are found to be

\frac{{1}}{{2}}

\frac{{1}}{{\sqrt{{2}}}}

, and

\frac{{1}}{{2}}

, and using the fact that the plane is normal to

\vec{{v}}

, the equation

a+\sqrt{{2}}b+c=1

is derived.

More Questions from DOT PRODUCT OF VECTORS

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

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\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