JEE MATHEMATICS MCQ: ADDITION OF VECTORS

1 Easy

1 of 1

Application +4 | -1

Let

P(3,2,6)

be a point in space and

Q

be a point on the line

\mathbf{r}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}})

. Then the value of

\mu

for which the vector

P Q

is parallel to the plane

x-4 y+3 z=1

View answer and explanation

Correct option: A

\frac{1}{4}

Option A (

\frac{{1}}{{4}}

) is correct because it represents the value of

\mu

for which the vector

PQ

is parallel to the given plane, calculated by setting the dot product of the normal vector of the plane and the vector

PQ

to zero.

More Questions from ADDITION OF VECTORS

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\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}

\mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}

\mathbf{d}

a

\mathbf{a} \cdot \mathbf{d}=0=(b \mathbf{c}

\mathbf{d}

{\mathbf{\vec a}}

2\;p

{\mathbf{\vec a}}

p + 1

\vec{a}, \vec{b}

\vec{c}

\vec{d}

\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}

\overrightarrow{\mathrm{c}}

|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|^2

\vec{a}, \vec{b}

\vec{c}

\vec{a}=8 \vec{b}

\vec{c}=-7 \vec{b}

\vec{a}

\vec{c}

2 \mathbf{i}+\mathbf{j}-\mathbf{k},-\mathbf{i}+2 \mathbf{j}+\lambda \mathbf{k}

-5 \mathbf{i}+2 \mathbf{j}-\mathbf{k}

\lambda

AD

ABC

E,\;BE

AC

F

AF:AC =