1

JEE Main 2019 (Online) 10th January Morning Slot

Let A be a point on the line $\overrightarrow r = \left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$ and B(3, 2, 6) be a point in the space. Then the value of $\mu$ for which the vector $\overrightarrow {AB}$  is parallel to the plane x $-$ 4y + 3z = 1 is -
A
${1 \over 8}$
B
${1 \over 2}$
C
${1 \over 4}$
D
$-$ ${1 \over 4}$

Explanation

Let point A is

$\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$

and point B is (3, 2, 6)

then $\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\widehat k$

which is parallel to the the plane x $-$ 4y + 3z = 1

$\therefore$  2 + 3$\mu$ $-$ 12 + 4$\mu$ + 12 $-$ 15$\mu$ = 0

8$\mu$ = 2

$\mu$ = ${1 \over 4}$
2

JEE Main 2019 (Online) 10th January Evening Slot

On which of the following lines lies the point of intersection of the line,   ${{x - 4} \over 2} = {{y - 5} \over 2} = {{z - 3} \over 1}$  and the plane, x + y + z = 2 ?
A
${{x - 4} \over 1} = {{y - 5} \over 1} = {{z - 5} \over { - 1}}$
B
${{x - 2} \over 2} = {{y - 3} \over 2} = {{z + 3} \over 3}$
C
${{x - 1} \over 1} = {{y - 3} \over 2} = {{z + 4} \over { - 5}}$
D
${{x + 3} \over 3} = {{4 - y} \over 3} = {{z + 1} \over { - 2}}$

Explanation

General point on the given line is

x = 2$\lambda$ + 4

y = 2$\lambda$ + 5

z = $\lambda$ + 3

Solving with plane,

2$\lambda$ + 4 + 2$\lambda$ + 5 + $\lambda$ + 3 = 2

5$\lambda$ + 12 = 2

5$\lambda$ = $-$ 10

$\lambda$ = $-$ 2
3

JEE Main 2019 (Online) 10th January Evening Slot

If $\overrightarrow \alpha$ = $\left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b$  and  $\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b$ be two given vectors $\overrightarrow a$ and $\overrightarrow b$ are non-collinear. The value of $\lambda$ for which vectors $\overrightarrow \alpha$ and $\overrightarrow \beta$ are collinear, is -
A
4
B
3
C
$-$3
D
$-$4

Explanation

$\overrightarrow \alpha = \left( {\lambda - 2} \right)\overrightarrow \alpha + \overrightarrow b$

$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow \alpha + 3\overrightarrow b$

${{\lambda - 2} \over {4\lambda - 2}} = {1 \over 3}$

$3\lambda - 6 = 4\lambda - 2$

$\lambda = - 4$
4

JEE Main 2019 (Online) 10th January Evening Slot

The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ?
A
(2, 1, 3)
B
(4, $-$ 1, 2)
C
(4, 1, $-$ 2)
D
($-$ 2, 3, 5)

Explanation

p : 3(x $-$ 0) + 5(y $-$ 2) + 1 (z $-$ 5) = 0

3x + 5y + z = 15