3D Geometry
436 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations l + m $-$ n = 0 and l2 + m2 $-$ n2 = 0. Then the value of sin4$\alpha$ + cos4$\alpha$ is :
A.
${{3 \over 8}}$
B.
${{3 \over 4}}$
C.
${{1 \over 2}}$
D.
${{5 \over 8}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
Let a, b$ \in $R. If the mirror image of the point P(a, 6, 9) with respect to the line
${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$ is (20, b, $-$a$-$9), then | a + b |, is equal to :
${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$ is (20, b, $-$a$-$9), then | a + b |, is equal to :
A.
88
B.
90
C.
86
D.
84
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
The vector equation of the plane passing through the intersection
of the planes $\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$ and $\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$, and the point (1, 0, 2) is :
of the planes $\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$ and $\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$, and the point (1, 0, 2) is :
A.
$\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = {7 \over 3}$
B.
$\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = 7$
C.
$\overrightarrow r .\left( {3\widehat i + 7\widehat j + 3\widehat k} \right) = 7$
D.
$\overrightarrow r .\left( {\widehat i - 7\widehat j + 3\widehat k} \right) = {7 \over 3}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
The equation of the plane passing through the point (1, 2, -3) and perpendicular to the
planes
3x + y - 2z = 5 and 2x - 5y - z = 7, is :
3x + y - 2z = 5 and 2x - 5y - z = 7, is :
A.
6x - 5y + 2z + 10 =0
B.
3x - 10y - 2z + 11 = 0
C.
6x - 5y - 2z - 2 = 0
D.
11x + y + 17z + 38 = 0
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
The distance of the point (1, 1, 9) from the point of intersection of the line
${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$
and the plane x + y + z = 17 is :
A.
$19\sqrt 2 $
B.
$2\sqrt {19} $
C.
38
D.
$\sqrt {38} $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
Suppose, the line ${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$ lies on the plane $x + 3y - 2z + \beta = 0$. Then $(\alpha + \beta )$ is equal to _______.
Correct Answer: 7
Explanation:
Given equation of line
${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$ ...... (i)
and plane x + 3y $-$ 2z + $\beta$ = 0 ...... (ii)
Line (i) passes through (2, 2, $-$2)
which lies on plane (ii).
$\therefore$ 2 + 6 + 4 + $\beta$ = 0 $\Rightarrow$ $\beta$ = $-$ 12
Also, given line is perpendicular to normal of the plane
$\alpha$(1) $-$ 5(3) + 2($-$2) = 0 $\Rightarrow$ $\alpha$ = 19
$\therefore$ $\alpha$ + $\beta$ = 19 + (-12) = 19 - 12 = 7
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
The square of the distance of the point of intersection
of the line ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$ and the plane $2x - y + z = 6$ from the point ($-$1, $-$1, 2) is __________.
of the line ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6}$ and the plane $2x - y + z = 6$ from the point ($-$1, $-$1, 2) is __________.
Correct Answer: 61
Explanation:
${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6} = \lambda $
$x = 2\lambda + 1,y = 3\lambda + 2,z = 6\lambda - 1$
for point of intersection of line & plane
$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$
$7\lambda = 7 \Rightarrow \lambda = 1$
point : (3, 5, 5)
(distance)2 = ${(3 + 1)^2} + {(5 + 1)^2} + {(5 - 2)^2}$
$ = 16 + 36 + 9 = 61$
$x = 2\lambda + 1,y = 3\lambda + 2,z = 6\lambda - 1$
for point of intersection of line & plane
$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$
$7\lambda = 7 \Rightarrow \lambda = 1$
point : (3, 5, 5)
(distance)2 = ${(3 + 1)^2} + {(5 + 1)^2} + {(5 - 2)^2}$
$ = 16 + 36 + 9 = 61$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x $-$ y + z + 3 = 0 and let R(3, 5, $\gamma$) be a point of this plane. Then the square of the length of the line segment SR is ___________.
Correct Answer: 72
Explanation:
Since R(3, 5, $\gamma$) lies on the plane 2x $-$ y + z + 3 = 0.
Therefore, 6 $-$ 5 + $\gamma$ + 3 = 0
$\Rightarrow$ $\gamma$ = $-$4
Now,
dr's of line QS are 2, $-$1, 1
equation of line QS is
${{x - 1} \over 2} = {{y - 3} \over { - 1}} = {{z - 4} \over 1} = \lambda $ (say)
$ \Rightarrow F(2\lambda + 1, - \lambda + 3,\lambda + 4)$
F lies in the plane
$ \Rightarrow 2(2\lambda + 1) - ( - \lambda + 3) + (\lambda + 4) + 3$ = 0
$ \Rightarrow 4\lambda + 2 + \lambda - 3 + \lambda + 7 = 0$
$ \Rightarrow 6\lambda + 6 = 0 \Rightarrow \lambda = - 1$
$\Rightarrow$ F($-$1, 4, 3)
Since, F is mid-point of QS.
Therefore, coordinated of S are ($-$3, 5, 2).
So, SR = $\sqrt {36 + 0 + 36} = \sqrt {72} $
SR2 = 72.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
Let Q be the foot of the perpendicular from the point P(7, $-$2, 13) on the plane containing the lines ${{x + 1} \over 6} = {{y - 1} \over 7} = {{z - 3} \over 8}$ and ${{x - 1} \over 3} = {{y - 2} \over 5} = {{z - 3} \over 7}$. Then (PQ)2, is equal to ___________.
Correct Answer: 96
Explanation:
Containing the line $\left| {\matrix{
{x + 1} & {y - 1} & {z - 3} \cr
6 & 7 & 8 \cr
3 & 5 & 7 \cr
} } \right| = 0$
$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$
$x - 2y + z = 0$
$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6 $
$P{Q^2} = 96$
$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$
$x - 2y + z = 0$
$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6 $
$P{Q^2} = 96$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
Let the line L be the projection of the line ${{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 4} \over 2}$ in the plane x $-$ 2y $-$ z = 3. If d is the distance of the point (0, 0, 6) from L, then d2 is equal to _______________.
Correct Answer: 26
Explanation:
To find the projection let's find the foot of perpendicular from $(1,3$,
4) to plane $x-2 y-z=3$
$ \begin{aligned} & \frac{x-1}{1}=\frac{y-3}{-2}=\frac{z-4}{-1}=\lambda_1 \\\\ & \left(\lambda_1+1\right)-2\left(-2 \lambda_1+3\right)-\left(-\lambda_1+4\right)=3 \\\\ & \Rightarrow 6 \lambda_1=12 \Rightarrow \lambda_1=2 \end{aligned} $
So, foot of perpendicular from $(1,3,4)$ to plane $x-2 y-z=3$ is $A$ $(3,-1,2)$.
Let us also find the intersection point of the plane and line
$ \begin{gathered} \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}=\lambda_2 \\\\ \left(2 \lambda_2+1\right)-2\left(\lambda_2+3\right)-\left(2 \lambda_2+4\right)=3-2 \lambda_2=12 \Rightarrow \lambda_2=-6 \end{gathered} $
The intersection point of the plane and line is $B(-11,-3,-8)$ Line passing through $A$ and $B$ is
$ \begin{aligned} & \frac{x-3}{-14}=\frac{y+1}{-2}=\frac{z-2}{-10}=\mu \\\\ & \frac{x-3}{7}=\frac{y+1}{1}=\frac{z-2}{5}=\mu \end{aligned} $
Now, let's find the distance from $O(0,0,6)$ to this line $L$.
Let's say $C(7 \mu+3, \mu-1,5 \mu+2)$ is any point on $L$. Then,
$ \begin{aligned} & \{(7 \mu+3)-0\} \cdot 7+\{(\mu-1)-0\} \cdot 1+\{(5 \mu+2)-6\} \cdot 5=0 \\\\ & \Rightarrow 49 \mu+21+\mu-1+25 \mu-20=0 \Rightarrow \mu=0 \\\\ & \therefore C(3,-1,2) \\\\ & \text { Distance }=\sqrt{(3-0)^2+(-1-0)^2+(2-6)^2}=\sqrt{26} \\\\ & d^2=26 \end{aligned} $
$ \begin{aligned} & \frac{x-1}{1}=\frac{y-3}{-2}=\frac{z-4}{-1}=\lambda_1 \\\\ & \left(\lambda_1+1\right)-2\left(-2 \lambda_1+3\right)-\left(-\lambda_1+4\right)=3 \\\\ & \Rightarrow 6 \lambda_1=12 \Rightarrow \lambda_1=2 \end{aligned} $
So, foot of perpendicular from $(1,3,4)$ to plane $x-2 y-z=3$ is $A$ $(3,-1,2)$.
Let us also find the intersection point of the plane and line
$ \begin{gathered} \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}=\lambda_2 \\\\ \left(2 \lambda_2+1\right)-2\left(\lambda_2+3\right)-\left(2 \lambda_2+4\right)=3-2 \lambda_2=12 \Rightarrow \lambda_2=-6 \end{gathered} $
The intersection point of the plane and line is $B(-11,-3,-8)$ Line passing through $A$ and $B$ is
$ \begin{aligned} & \frac{x-3}{-14}=\frac{y+1}{-2}=\frac{z-2}{-10}=\mu \\\\ & \frac{x-3}{7}=\frac{y+1}{1}=\frac{z-2}{5}=\mu \end{aligned} $
Now, let's find the distance from $O(0,0,6)$ to this line $L$.
Let's say $C(7 \mu+3, \mu-1,5 \mu+2)$ is any point on $L$. Then,
$ \begin{aligned} & \{(7 \mu+3)-0\} \cdot 7+\{(\mu-1)-0\} \cdot 1+\{(5 \mu+2)-6\} \cdot 5=0 \\\\ & \Rightarrow 49 \mu+21+\mu-1+25 \mu-20=0 \Rightarrow \mu=0 \\\\ & \therefore C(3,-1,2) \\\\ & \text { Distance }=\sqrt{(3-0)^2+(-1-0)^2+(2-6)^2}=\sqrt{26} \\\\ & d^2=26 \end{aligned} $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points. Q(3, $-$4, $-$5) and R(2, $-$3, 1) and the plane 2x + y + z = 7, is equal to ______________.
Correct Answer: 7
Explanation:
$\overrightarrow {QR} : - {{x - 3} \over 1} = {{y + 4} \over { - 1}} = {{z + 5} \over { - 6}} = r$
$ \Rightarrow (x,y,z) \equiv (r + 3, - r - 4, - 6r - 5)$
Now, satisfying it in the given plane.
We get r = $-$2
so, required point of intersection is T(1, $-$2, 7).
Hence, PT = 7.
$ \Rightarrow (x,y,z) \equiv (r + 3, - r - 4, - 6r - 5)$
Now, satisfying it in the given plane.
We get r = $-$2
so, required point of intersection is T(1, $-$2, 7).
Hence, PT = 7.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
Let a plane P pass through the point (3, 7, $-$7) and contain the line, ${{x - 2} \over { - 3}} = {{y - 3} \over 2} = {{z + 2} \over 1}$. If distance of the plane P from the origin is d, then d2 is equal to ______________.
Correct Answer: 3
Explanation:
$\overrightarrow {BA} = (\widehat i + 4\widehat j - 5\widehat k)$
$\overrightarrow {BA} \times \overrightarrow l = \overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 3} & 2 & 1 \cr 1 & 4 & { - 5} \cr } } \right|$
$a\widehat i + b\widehat j + c\widehat k = - 14\widehat i - \widehat j(14) + \widehat k( - 14)$
a = 1, b = 1, c = 1
Plane is (x $-$ 2) + (y $-$ 3) + (z + 2) = 0
$ \Rightarrow $ x + y + z $-$ 3 = 0
$ \therefore $ d = $\sqrt 3 $ $\Rightarrow$ d2 = 3
$\overrightarrow {BA} \times \overrightarrow l = \overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 3} & 2 & 1 \cr 1 & 4 & { - 5} \cr } } \right|$
$a\widehat i + b\widehat j + c\widehat k = - 14\widehat i - \widehat j(14) + \widehat k( - 14)$
a = 1, b = 1, c = 1
Plane is (x $-$ 2) + (y $-$ 3) + (z + 2) = 0
$ \Rightarrow $ x + y + z $-$ 3 = 0
$ \therefore $ d = $\sqrt 3 $ $\Rightarrow$ d2 = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
If the lines ${{x - k} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$ and
${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$ are co-planar, then the value of k is _____________.
${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$ are co-planar, then the value of k is _____________.
Correct Answer: 1
Explanation:
$\left| {\matrix{
{k + 1} & 4 & 6 \cr
1 & 2 & 3 \cr
3 & 2 & 1 \cr
} } \right| = 0$
$ \Rightarrow $ $(k + 1)[2 - 6] - 4[1 - 9] + 6[2 - 6] = 0$
$ \Rightarrow $ $k = 1$
$ \Rightarrow $ $(k + 1)[2 - 6] - 4[1 - 9] + 6[2 - 6] = 0$
$ \Rightarrow $ $k = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
Let P be a plane passing through the points (1, 0, 1), (1, $-$2, 1) and (0, 1, $-$2). Let a vector $\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ be such that $\overrightarrow a $ is parallel to the plane P, perpendicular to $(\widehat i + 2\widehat j + 3\widehat k)$ and $\overrightarrow a \,.\,(\widehat i + \widehat j + 2\widehat k) = 2$, then ${(\alpha - \beta + \gamma )^2}$ equals ____________.
Correct Answer: 81
Explanation:
Equation of plane :
$\left| {\matrix{ {x - 1} & {y - 0} & {z - 1} \cr {1 - 1} & 2 & {1 - 1} \cr {1 - 0} & {0 - 1} & {1 + 2} \cr } } \right| = 0$
$ \Rightarrow 3x - z - 2 = 0$
$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ || to 3x $-$ z $-$ 2 = 0
$ \Rightarrow 3\alpha - 8 = 0$ ..... (1)
$\overrightarrow a \bot \widehat i + \widehat j + 3\widehat k$
$ \Rightarrow \alpha + 2\beta + 3\gamma = 0$ ...... (2)
$\overrightarrow a .(\widehat i + \widehat j + 2\widehat k) = 0$
$\Rightarrow$ $\alpha$ + $\beta$ + 2$\gamma$ = 2 ........ (3)
On solving 1, 2 & 3
$\alpha$ = 1, $\beta$ = $-$5, $\gamma$ = 3
So, ($\alpha$ $-$ $\beta$ + $\gamma$)2 = 81
$\left| {\matrix{ {x - 1} & {y - 0} & {z - 1} \cr {1 - 1} & 2 & {1 - 1} \cr {1 - 0} & {0 - 1} & {1 + 2} \cr } } \right| = 0$
$ \Rightarrow 3x - z - 2 = 0$
$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ || to 3x $-$ z $-$ 2 = 0
$ \Rightarrow 3\alpha - 8 = 0$ ..... (1)
$\overrightarrow a \bot \widehat i + \widehat j + 3\widehat k$
$ \Rightarrow \alpha + 2\beta + 3\gamma = 0$ ...... (2)
$\overrightarrow a .(\widehat i + \widehat j + 2\widehat k) = 0$
$\Rightarrow$ $\alpha$ + $\beta$ + 2$\gamma$ = 2 ........ (3)
On solving 1, 2 & 3
$\alpha$ = 1, $\beta$ = $-$5, $\gamma$ = 3
So, ($\alpha$ $-$ $\beta$ + $\gamma$)2 = 81
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
Let the mirror image of the point (1, 3, a) with respect to the plane $\overrightarrow r .\left( {2\widehat i - \widehat j + \widehat k} \right) - b = 0$ be ($-$3, 5, 2). Then, the value of | a + b | is equal to ____________.
Correct Answer: 1
Explanation:
Given equation of plane in vector form is $\overrightarrow r \,.\,(2\widehat i - \widehat j + \widehat k) - b = 0$
Its Cartesian form will be
$2x - y + z = b$ ...... (i)
$\because$ R is the mid-point of PQ.
$\therefore$ $R \equiv {{P + Q} \over 2} \Rightarrow R \equiv \left( { - 1,4,{{a + 2} \over 2}} \right)$
$\because$ R lies on the plane (i).
$\therefore$ $ - 2 - 4 + {{a + 2} \over 2} = b \Rightarrow a + 2 = 2b + 12$
$ \Rightarrow a = 2b + 10$ ....... (ii)
$\because$ Direction ratio's of QP is $(1 - ( - 3),3 - 5,a - 2)$
i.e. $(4, - 2,a - 2)$
and direction ratios of normal to the given plane are (2, $-$1, 1)
$\because$ n and QP are parallel.
$\therefore$ ${2 \over 4} = {{ - 1} \over { - 2}} = {1 \over {a - 2}}$
$\therefore$ $a - 2 = 2 \Rightarrow a = 4$
From Eq. (ii), b = $-$3
$\therefore$ $|a + b| = |4 - 3| = |1| = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
Let P be a plane containing the line ${{x - 1} \over 3} = {{y + 6} \over 4} = {{z + 5} \over 2}$ and parallel to the line ${{x - 1} \over 4} = {{y - 2} \over { - 3}} = {{z + 5} \over 7}$. If the point (1, $-$1, $\alpha$) lies on the plane P, then the value of |5$\alpha$| is equal to ____________.
Correct Answer: 38
Explanation:
Equation of required plane is $\left| {\matrix{ {x - 1} & {y + 6} & {z + 5} \cr 3 & 4 & 2 \cr 4 & { - 3} & 7 \cr } } \right| = 0$
Since, (1, $-$1, $\alpha$) lies on it,
So, replace x by 1, y by ($-$1) and z and $\alpha$.
$\left| {\matrix{ 0 & 5 & {\alpha + 5} \cr 3 & 4 & 2 \cr 4 & { - 3} & 7 \cr } } \right| = 0$
$ \Rightarrow 5\alpha + 38 = 0 \Rightarrow 5\alpha = - 38$
$\therefore$ $\left| {5\alpha } \right| = \left| { - 38} \right| = 38$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
Let the plane ax + by + cz + d = 0 bisect the line joining the points (4, $-$3, 1) and (2, 3, $-$5) at the right angles. If a, b, c, d are integers, then the
minimum value of (a2 + b2 + c2 + d2) is _________.
minimum value of (a2 + b2 + c2 + d2) is _________.
Correct Answer: 28
Explanation:
Normal of plane = $\overrightarrow {PQ} = - 2\widehat i + 6\widehat j - 6\widehat k$
a = $-$2, b = 6, c = $-$6
& equation of plane is
$-$2x + 6y $-$ 6z + d = 0
$ M(3,0, - 2)$ is the midpoint of the line which present on the plane
which satisfy the plane
$ \therefore $ d = $-$6
Now equation of plane is
$-$2x + 6y $-$ 6z $-$ 6 = 0
x $-$ 3y + 3z + 3 = 0
$ \Rightarrow $ (a2 + b2 + c2 + d2)min = 12 + 9 + 9 + 9 = 28
a = $-$2, b = 6, c = $-$6
& equation of plane is
$-$2x + 6y $-$ 6z + d = 0
$ M(3,0, - 2)$ is the midpoint of the line which present on the plane
which satisfy the plane
$ \therefore $ d = $-$6
Now equation of plane is
$-$2x + 6y $-$ 6z $-$ 6 = 0
x $-$ 3y + 3z + 3 = 0
$ \Rightarrow $ (a2 + b2 + c2 + d2)min = 12 + 9 + 9 + 9 = 28
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
The equation of the planes parallel to the plane x $-$ 2y + 2z $-$ 3 = 0 which are at unit distance from the point (1, 2, 3) is ax + by + cz + d = 0. If (b $-$ d) = k(c $-$ a), then the positive value of k is :
Correct Answer: 4
Explanation:
The equation of the planes parallel to the plane x $-$ 2y + 2z $-$ 3 = 0
$x - 2y + 2z + \lambda = 0$
Now given
$d = {{\left| {1 - 4 + 6 + \lambda } \right|} \over {\sqrt 9 }} = 1$
$\left| {\lambda + 3} \right| = 3$
$\lambda + 3 = \pm 3 \Rightarrow \lambda = 0, - 6$
So planes are : $x - 2y + 2z - 6 = 0$
and $x - 2y + 2z = 0$
$b - d = - 2 + 6 = 4$
$c - a = 2 - 1 = 1$
$ \therefore $ $ {{b - d} \over {c - a}} = k$
$ \Rightarrow k = 4$
$x - 2y + 2z + \lambda = 0$
Now given
$d = {{\left| {1 - 4 + 6 + \lambda } \right|} \over {\sqrt 9 }} = 1$
$\left| {\lambda + 3} \right| = 3$
$\lambda + 3 = \pm 3 \Rightarrow \lambda = 0, - 6$
So planes are : $x - 2y + 2z - 6 = 0$
and $x - 2y + 2z = 0$
$b - d = - 2 + 6 = 4$
$c - a = 2 - 1 = 1$
$ \therefore $ $ {{b - d} \over {c - a}} = k$
$ \Rightarrow k = 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
Let P be an arbitrary point having sum of the squares of the distances from the planes x + y + z = 0, lx $-$ nz = 0 and x $-$ 2y + z = 0, equal to 9. If the locus of the point P is x2 + y2 + z2 = 9, then the value of l $-$ n is equal to _________.
Correct Answer: 0
Explanation:
Let point P is ($\alpha$, $\beta$, $\gamma$)
${\left( {{{\alpha + \beta + \gamma } \over {\sqrt 3 }}} \right)^2} + {\left( {{{l\alpha - n\gamma } \over {\sqrt {{l^2} + {n^2}} }}} \right)^2} + {\left( {{{\alpha - 2\beta + \gamma } \over {\sqrt 6 }}} \right)^2} = 9$
Locus is ${{{{(x + y + z)}^2}} \over 3} + {{{{(\ln - nz)}^2}} \over {{l^2} + {n^2}}} + {{{{(x - 2y + z)}^2}} \over 6} = 9$
${x^2}\left( {{1 \over 2} + {{{l^2}} \over {{l^2} + {n^2}}}} \right) + {y^2} + {z^2}\left( {{1 \over 2} + {{{n^2}} \over {{l^2} + {n^2}}}} \right) + 2zx\left( {{1 \over 2} - {{\ln } \over {{l^2} + {n^2}}}} \right) - 9 = 0$
Since its given that ${x^2} + {y^2} + {z^2} = 9$
After solving l = n,
then, l $-$ n = 0
${\left( {{{\alpha + \beta + \gamma } \over {\sqrt 3 }}} \right)^2} + {\left( {{{l\alpha - n\gamma } \over {\sqrt {{l^2} + {n^2}} }}} \right)^2} + {\left( {{{\alpha - 2\beta + \gamma } \over {\sqrt 6 }}} \right)^2} = 9$
Locus is ${{{{(x + y + z)}^2}} \over 3} + {{{{(\ln - nz)}^2}} \over {{l^2} + {n^2}}} + {{{{(x - 2y + z)}^2}} \over 6} = 9$
${x^2}\left( {{1 \over 2} + {{{l^2}} \over {{l^2} + {n^2}}}} \right) + {y^2} + {z^2}\left( {{1 \over 2} + {{{n^2}} \over {{l^2} + {n^2}}}} \right) + 2zx\left( {{1 \over 2} - {{\ln } \over {{l^2} + {n^2}}}} \right) - 9 = 0$
Since its given that ${x^2} + {y^2} + {z^2} = 9$
After solving l = n,
then, l $-$ n = 0
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
If the equation of the plane passing through the line of intersection of the planes 2x $-$ 7y + 4z $-$ 3 = 0, 3x $-$ 5y + 4z + 11 = 0 and the point ($-$2, 1, 3) is ax + by + cz $-$ 7 = 0, then the value of 2a + b + c $-$ 7 is ____________.
Correct Answer: 4
Explanation:
Equation of plane can be written using family of planes : P1 + $\lambda$P2 = 0
(2x $-$ 7y + 4z $-$ 3) + $\lambda$ (3x $-$ 5y + 4z + 11) = 0
It passes through ($-$2, 1, 3)
$ \therefore $ ($-$4 + 7 + 12 $-$ 3) + $\lambda$ ($-$6 $-$ 5 + 12 + 11) = 0
$-$2 + $\lambda$ (12) = 0
$\lambda$ = ${1 \over 6}$.
$ \therefore $ 12x $-$ 42y + 24z $-$ 18 + 3x $-$ 5y + 4z + 11 = 0
15x $-$ 47y + 28z $-$ 7 = 0
$ \therefore $ a = 15, b = $-$47, c = 28
$ \therefore $ 2a + b + c $-$ 7 = 30 $-$ 47 + 28 $-$ 7 = 4
(2x $-$ 7y + 4z $-$ 3) + $\lambda$ (3x $-$ 5y + 4z + 11) = 0
It passes through ($-$2, 1, 3)
$ \therefore $ ($-$4 + 7 + 12 $-$ 3) + $\lambda$ ($-$6 $-$ 5 + 12 + 11) = 0
$-$2 + $\lambda$ (12) = 0
$\lambda$ = ${1 \over 6}$.
$ \therefore $ 12x $-$ 42y + 24z $-$ 18 + 3x $-$ 5y + 4z + 11 = 0
15x $-$ 47y + 28z $-$ 7 = 0
$ \therefore $ a = 15, b = $-$47, c = 28
$ \therefore $ 2a + b + c $-$ 7 = 30 $-$ 47 + 28 $-$ 7 = 4
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
If the distance of the point (1, $-$2, 3) from the plane x + 2y $-$ 3z + 10 = 0 measured parallel to the line, ${{x - 1} \over 3} = {{2 - y} \over m} = {{z + 3} \over 1}$ is $\sqrt {{7 \over 2}} $, then the value of |m| is equal to _________.
Correct Answer: 2
Explanation:
Given line L,
${{x - 1} \over 3} = {{2 - y} \over m} = {{z + 3} \over 1}$
$ \Rightarrow $ ${{x - 1} \over 3} = {{y - 2} \over -m} = {{z + 3} \over 1}$
$ \therefore $ D.R of line = <3, -m, 1>
D.R of parallel line PQ will also be same.
$ \therefore $ Equation of line PQ,
${{x - 1} \over 3} = {{y + 2} \over { - m}} = {{z - 3} \over 1} = \lambda $
Pt. $Q(3\lambda + 1, - m\lambda - 2,\lambda + 3)$ lie on plane
$(3\lambda + 1) + 2( - m\lambda - 2) - 3(\lambda + 3) + 10 = 0$
$ \Rightarrow 3\lambda - 2m\lambda - 3\lambda + 1 - 4 - 9 + 10 = 0$
$ \Rightarrow - 2m\lambda = 2$
$m\lambda = - 1 \Rightarrow \lambda = - {1 \over m}$
$Q\left[ { - {3 \over m} + 1, - 1, - {1 \over m} + 3} \right]$
Given, $PQ = \sqrt {{7 \over 2}} $
$ \Rightarrow $ $\sqrt {{{\left( { - {3 \over m}} \right)}^2} + 1 + {{\left( { - {1 \over m}} \right)}^2}} = \sqrt {{7 \over 2}} $
$ \Rightarrow {{10 + {m^2}} \over {{m^2}}} = {7 \over 2}$
$ \Rightarrow 20 + 2{m^2} = 7{m^2}$
$ \Rightarrow $ ${m^2} = 4 \Rightarrow |m| = 2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
Let ($\lambda$, 2, 1) be a point on the plane which passes through the point (4, $-$2, 2). If the plane is perpendicular to the line joining the points ($-$2, $-$21, 29) and ($-$1, $-$16, 23), then ${\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$ is equal to __________.
Correct Answer: 8
Explanation:
$\overrightarrow {AB} \bot \overrightarrow {PQ} $
$\left[ {(4 - \lambda )\widehat i - 4\widehat j + \widehat k} \right].\left[ { + \widehat i + 5\widehat j - 6\widehat k} \right] = 0$
$4 - \lambda - 20 - 6 = 0$
$ \Rightarrow $ $\lambda $ = -22
Now, ${\lambda \over {11}} = - 2$
$ \Rightarrow {\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$
$ \Rightarrow 4 + 8 - 4 = 8$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
A line 'l' passing through origin is perpendicular to the lines
${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$
${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$
If the co-ordinates of the point in the first octant on 'l2‘ at a distance of $\sqrt {17} $ from the point of intersection of 'l' and 'l1' are (a, b, c) then 18(a + b + c) is equal to ___________.
${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$
${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$
If the co-ordinates of the point in the first octant on 'l2‘ at a distance of $\sqrt {17} $ from the point of intersection of 'l' and 'l1' are (a, b, c) then 18(a + b + c) is equal to ___________.
Correct Answer: 44
Explanation:
${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$
${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $ D.R. of ${l_1} = 1,2,2$
${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$
${l_2}:{{x - 3} \over 2} = {{y - 3} \over 2} = {{z - 2} \over 1} \Rightarrow $ D.R. of ${l_2} = 2,2,1$
D.R. of l is $ \bot $ to l1 & k2
$ \therefore $ D.R. of $l\,||\,({l_1} \times {l_2}) \Rightarrow ( - 2,3 - 2)$
$ \therefore $ Equation of $l:{x \over 2} = {y \over { - 3}} = {z \over 2}$
Solving l & l1
$(2\lambda , - 3\lambda ,2\lambda ) = (\mu + 3,2\mu - 1,2\mu + \mu )$
$ \Rightarrow 2\lambda = \mu + 3$
$ - 3\lambda = 2\mu - 1$
$2\lambda = 2\mu + 4$
$ \Rightarrow \mu + 3 = 2\mu + 4$
$\mu = - 1$
$\lambda = 1$
$P(2, - 3,2)$ {intersection point}
Let, $Q(2v + 3,2v + 3,v + 2)$ be point on l2
Now, $PQ = \sqrt {{{(2v + 3 - 2)}^2} + {{(2v + 3 + 3)}^2} + {{(v + 2 - 2)}^2}} = \sqrt {17} $
$ \Rightarrow {(2v + 1)^2} + {(2v + 6)^2} + {(v)^2} = 17$
$ \Rightarrow 9{v^2} + 28v + 36 + 1 - 17 = 0$
$ \Rightarrow 9{v^2} + 28v + 20 = 0$
$ \Rightarrow 9{v^2} + 18v + 10v + 20 = 0$
$ \Rightarrow (9v + 10)(v + 2) = 0$
$ \Rightarrow v = - 2$ (rejected), $ - {{10} \over 9}$ (accepted)
$Q\left( {3 - {{20} \over 9},3 - {{20} \over 9},2 - {{10} \over 9}} \right)$
$\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$
$ \therefore $ $18(a + b + c)$
$ = 18\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$
$ = 44$
${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $ D.R. of ${l_1} = 1,2,2$
${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$
${l_2}:{{x - 3} \over 2} = {{y - 3} \over 2} = {{z - 2} \over 1} \Rightarrow $ D.R. of ${l_2} = 2,2,1$
D.R. of l is $ \bot $ to l1 & k2
$ \therefore $ D.R. of $l\,||\,({l_1} \times {l_2}) \Rightarrow ( - 2,3 - 2)$
$ \therefore $ Equation of $l:{x \over 2} = {y \over { - 3}} = {z \over 2}$
Solving l & l1
$(2\lambda , - 3\lambda ,2\lambda ) = (\mu + 3,2\mu - 1,2\mu + \mu )$
$ \Rightarrow 2\lambda = \mu + 3$
$ - 3\lambda = 2\mu - 1$
$2\lambda = 2\mu + 4$
$ \Rightarrow \mu + 3 = 2\mu + 4$
$\mu = - 1$
$\lambda = 1$
$P(2, - 3,2)$ {intersection point}
Let, $Q(2v + 3,2v + 3,v + 2)$ be point on l2
Now, $PQ = \sqrt {{{(2v + 3 - 2)}^2} + {{(2v + 3 + 3)}^2} + {{(v + 2 - 2)}^2}} = \sqrt {17} $
$ \Rightarrow {(2v + 1)^2} + {(2v + 6)^2} + {(v)^2} = 17$
$ \Rightarrow 9{v^2} + 28v + 36 + 1 - 17 = 0$
$ \Rightarrow 9{v^2} + 28v + 20 = 0$
$ \Rightarrow 9{v^2} + 18v + 10v + 20 = 0$
$ \Rightarrow (9v + 10)(v + 2) = 0$
$ \Rightarrow v = - 2$ (rejected), $ - {{10} \over 9}$ (accepted)
$Q\left( {3 - {{20} \over 9},3 - {{20} \over 9},2 - {{10} \over 9}} \right)$
$\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$
$ \therefore $ $18(a + b + c)$
$ = 18\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$
$ = 44$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
Let $\lambda$ be an integer. If the shortest distance between the lines
x $-$ $\lambda$ = 2y $-$ 1 = $-$2z and x = y + 2$\lambda$ = z $-$ $\lambda$ is ${{\sqrt 7 } \over {2\sqrt 2 }}$, then the value of | $\lambda$ | is _________.
x $-$ $\lambda$ = 2y $-$ 1 = $-$2z and x = y + 2$\lambda$ = z $-$ $\lambda$ is ${{\sqrt 7 } \over {2\sqrt 2 }}$, then the value of | $\lambda$ | is _________.
Correct Answer: 1
Explanation:
${{x - \lambda } \over 1} = {{y - {1 \over 2}} \over {{1 \over 2}}} = {z \over { - {1 \over 2}}}$
${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$ ....... (1)
Point on line = $\left( {\lambda ,{1 \over 2},0} \right)$
${x \over 1} = {{y + 2\lambda } \over 1} = {{z - \lambda } \over 1}$ ....... (2)
Point on line = $(0, - 2\lambda ,\lambda )$
Distance between skew lines $ = {{\left[ {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}{{\overrightarrow b }_1}{{\overrightarrow b }_2}} \right]} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}$
$\left| {\matrix{ \lambda & {{1 \over 2} + 2\lambda } & { - \lambda } \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|$
$\overline {\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|} $
$ = {{\left| { - 5\lambda - {3 \over 2}} \right|} \over {\sqrt {14} }} = {{\sqrt 7 } \over {2\sqrt 2 }}$
$ = |10\lambda + 3| = 7 \Rightarrow \lambda = - 1$
$ \Rightarrow |\lambda | = 1$
${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$ ....... (1)
Point on line = $\left( {\lambda ,{1 \over 2},0} \right)$
${x \over 1} = {{y + 2\lambda } \over 1} = {{z - \lambda } \over 1}$ ....... (2)
Point on line = $(0, - 2\lambda ,\lambda )$
Distance between skew lines $ = {{\left[ {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}{{\overrightarrow b }_1}{{\overrightarrow b }_2}} \right]} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}$
$\left| {\matrix{ \lambda & {{1 \over 2} + 2\lambda } & { - \lambda } \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|$
$\overline {\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|} $
$ = {{\left| { - 5\lambda - {3 \over 2}} \right|} \over {\sqrt {14} }} = {{\sqrt 7 } \over {2\sqrt 2 }}$
$ = |10\lambda + 3| = 7 \Rightarrow \lambda = - 1$
$ \Rightarrow |\lambda | = 1$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
A plane P meets the coordinate axes at A, B
and C respectively. The centroid of $\Delta $ABC is
given to be (1, 1, 2). Then the equation of the
line through this centroid and perpendicular to
the plane P is :
A.
${{x - 1} \over 1} = {{y - 1} \over 1} = {{z - 2} \over 2}$
B.
${{x - 1} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$
C.
${{x - 1} \over 2} = {{y - 1} \over 2} = {{z - 2} \over 1}$
D.
${{x - 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 2}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
The shortest distance between the lines
${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$
and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :
${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$
and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :
A.
1
B.
${1 \over 2}$
C.
${1 \over {\sqrt 2 }}$
D.
${1 \over {\sqrt 3 }}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
If for some $\alpha $ $ \in $ R, the lines
L1 : ${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$ and
L2 : ${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$ are coplanar,
then the line L2 passes through the point :
L1 : ${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$ and
L2 : ${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$ are coplanar,
then the line L2 passes through the point :
A.
(10, 2, 2)
B.
(2, –10, –2)
C.
(10, –2, –2)
D.
(–2, 10, 2)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
If (a, b, c) is the image of the point (1, 2, -3) in
the line ${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$, then a + b + c is :
the line ${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$, then a + b + c is :
A.
1
B.
2
C.
3
D.
-1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The distance of the point (1, –2, 3) from
the plane x – y + z = 5 measured parallel to
the line ${x \over 2} = {y \over 3} = {z \over { - 6}}$ is :
the plane x – y + z = 5 measured parallel to
the line ${x \over 2} = {y \over 3} = {z \over { - 6}}$ is :
A.
7
B.
1
C.
${1 \over 7}$
D.
${7 \over 5}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
The plane which bisects the line joining, the
points (4, –2, 3) and (2, 4, –1) at right angles
also passes through the point :
A.
(4, 0, 1)
B.
(0, –1, 1)
C.
(0, 1, –1)
D.
(4, 0, –1)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
The foot of the perpendicular drawn from the
point (4, 2, 3) to the line joining the points
(1, –2, 3) and (1, 1, 0) lies on the plane :
A.
x – 2y + z = 1
B.
x + 2y – z = 1
C.
x – y – 2y = 1
D.
2x + y – z = 1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
A plane passing through the point (3, 1, 1)
contains two lines whose direction ratios are 1,
–2, 2 and 2, 3, –1 respectively. If this plane also
passes through the point ($\alpha $, –3, 5), then
$\alpha $ is
equal to:
A.
-10
B.
10
C.
5
D.
-5
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
The plane passing through the points (1, 2, 1),
(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
also passes through the point :
(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
also passes through the point :
A.
(0, 6, –2)
B.
(–2, 0, 1)
C.
(0, –6, 2)
D.
(2, 0 –1)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
The mirror image of the point (1, 2, 3) in a plane
is
$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$. Which of the following points lies on this plane ?
$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$. Which of the following points lies on this plane ?
A.
(1, –1, 1)
B.
(–1, –1, –1)
C.
(–1, –1, 1)
D.
(1, 1, 1)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The shortest distance between the lines
${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$ and
${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$ is :
${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$ and
${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$ is :
A.
3
B.
${7 \over 2}\sqrt {30} $
C.
$3\sqrt {30} $
D.
$2\sqrt {30} $
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point
(2, 1, 6). Then the image of R in the plane P is :
A.
(4, 3, 2)
B.
(6, 5, - 2)
C.
(3, 4, -2)
D.
(6, 5, 2)
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
If the equation of a plane P, passing through the intersection of the planes,
x + 4y - z + 7 = 0 and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b $ \in $ R, then the distance of the point (3, 2, -1) from the plane P is...........
x + 4y - z + 7 = 0 and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b $ \in $ R, then the distance of the point (3, 2, -1) from the plane P is...........
Correct Answer: 3
Explanation:
Equation of plane P is
$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$
$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$
${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$
$ \therefore $ 15 - 75$\lambda $ = 42 - 48$\lambda $
$ \Rightarrow $ -27 = 27$\lambda $
$ \Rightarrow $ $\lambda $ = -1
$ \therefore $ Plane is $(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$
$ \Rightarrow $ $2x - 3y + 6z - 15 = 0$
Distance of (3, 2, -1) from the plane P
= ${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$
$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$
$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$
${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$
$ \therefore $ 15 - 75$\lambda $ = 42 - 48$\lambda $
$ \Rightarrow $ -27 = 27$\lambda $
$ \Rightarrow $ $\lambda $ = -1
$ \therefore $ Plane is $(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$
$ \Rightarrow $ $2x - 3y + 6z - 15 = 0$
Distance of (3, 2, -1) from the plane P
= ${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
Let a plane P contain two lines
$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$, $\lambda \in R$ and
$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$, $\mu \in R$
If Q($\alpha $, $\beta $, $\gamma $) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3($\alpha $ + $\beta $ + $\gamma $) equals _______.
$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$, $\lambda \in R$ and
$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$, $\mu \in R$
If Q($\alpha $, $\beta $, $\gamma $) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3($\alpha $ + $\beta $ + $\gamma $) equals _______.
Correct Answer: 5
Explanation:
Given lines,
$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ parallel to $(\widehat i + \widehat j)$
Let, $\overrightarrow {{n_1}} = (\widehat i + \widehat j)$
and $\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$ parallel to $(\widehat j - \widehat k)$
Let, $\overrightarrow {{n_2}} = (\widehat j - \widehat k)$
$ \therefore $ Normal of plane, $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $
$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & 0 \cr 0 & 1 & { - 1} \cr } } \right|$
$ = - \widehat i + \widehat j + \widehat k$
Line $\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.
$ \therefore $ Equation of the plane,
$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$
$ \Rightarrow x - y - z - 1 = 0$
Foot of perpendicular from (x1, y1, z1) on the plane,
${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$
Here foot of perpendicular is drawn from M(1, 0, 1),
$ \therefore $ ${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$
$ \therefore $ $x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$
${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$
${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$
According to the question,
$x = \alpha $, $y = \beta $, $z = \gamma $
$ \therefore $ $\alpha = {4 \over 3}$, $\beta = - {1 \over 3}$, $\gamma = {2 \over 3}$
$ \therefore $ $3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$
$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ parallel to $(\widehat i + \widehat j)$
Let, $\overrightarrow {{n_1}} = (\widehat i + \widehat j)$
and $\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$ parallel to $(\widehat j - \widehat k)$
Let, $\overrightarrow {{n_2}} = (\widehat j - \widehat k)$
$ \therefore $ Normal of plane, $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $
$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & 0 \cr 0 & 1 & { - 1} \cr } } \right|$
$ = - \widehat i + \widehat j + \widehat k$
Line $\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.
$ \therefore $ Equation of the plane,
$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$
$ \Rightarrow x - y - z - 1 = 0$
Foot of perpendicular from (x1, y1, z1) on the plane,
${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$
Here foot of perpendicular is drawn from M(1, 0, 1),
$ \therefore $ ${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$
$ \therefore $ $x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$
${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$
${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$
According to the question,
$x = \alpha $, $y = \beta $, $z = \gamma $
$ \therefore $ $\alpha = {4 \over 3}$, $\beta = - {1 \over 3}$, $\gamma = {2 \over 3}$
$ \therefore $ $3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
If the distance between the plane,
23x – 10y – 2z + 48 = 0 and the plane
containing the lines ${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$
and ${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$
is equal to ${k \over {\sqrt {633} }}$, then k is equal to ______.
containing the lines ${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$
and ${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$
is equal to ${k \over {\sqrt {633} }}$, then k is equal to ______.
Correct Answer: 3
Explanation:
Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)
$ \Rightarrow $ $\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ $\left| {{3 \over {\sqrt {633} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ k = 3
$ \Rightarrow $ $\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ $\left| {{3 \over {\sqrt {633} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ k = 3
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
The projection of the line segment joining the
points (1, –1, 3) and (2, –4, 11) on the line
joining the points (–1, 2, 3) and (3, –2, 10)
is ____________.
Correct Answer: 8
Explanation:
Let A (1, – 1, 3), B(2, – 4, 11), C (–1, 2, 3) & D (3, –2, 10)
$ \therefore $ $\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$
$ \Rightarrow $ $\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$
Projection of $\overrightarrow {AB} $ on $\overrightarrow {CD} $ = ${{\overrightarrow {AB} .\overrightarrow {CD} } \over {\left| {\overrightarrow {CD} } \right|}}$
= ${{4 + 12 + 56} \over {\sqrt {16 + 16 + 49} }}$
= ${{72} \over 9}$
= 8
$ \therefore $ $\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$
$ \Rightarrow $ $\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$
Projection of $\overrightarrow {AB} $ on $\overrightarrow {CD} $ = ${{\overrightarrow {AB} .\overrightarrow {CD} } \over {\left| {\overrightarrow {CD} } \right|}}$
= ${{4 + 12 + 56} \over {\sqrt {16 + 16 + 49} }}$
= ${{72} \over 9}$
= 8
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Evening Slot
If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through ($\alpha $, 7, 1)
is
$\left( {{5 \over 3},{7 \over 3},{{17} \over 3}} \right)$, then $\alpha $ is equal to______.
Correct Answer: 4
Explanation:
Direction Ratio of PQ are
= (${5 \over 3} - 1$, ${7 \over 3} - 0$, ${{17} \over 3} - 3$)
= (2, 7, 8)
Direction ratio of line QA are
= ($\alpha - {5 \over 3}$, $7 - {7 \over 3}$, 1 - ${{17} \over 3}$)
= (3$\alpha $ – 5, 14, –14)
PQ is perpendicular to line QA
$ \therefore $ $\overrightarrow {PQ} .\overrightarrow {QA} $ = 0
$ \Rightarrow $ 2(3$\alpha $ – 5) + 7.14 + (–14).8 = 0
$ \Rightarrow $ $\alpha $ = 4
2020
JEE Advanced
MSQ
JEE Advanced 2020 Paper 2 Offline
Let $\alpha $2 + $\beta $2 + $\gamma $2 $ \ne $ 0 and $\alpha $ + $\gamma $ = 1. Suppose the point (3, 2, $-$1) is the mirror image of the point (1, 0, $-$1) with respect to the plane $\alpha $x + $\beta $y + $\gamma $z = $\delta $. Then which of the following statements is/are TRUE?
A.
$\alpha $ + $\beta $ = 2
B.
$\delta $ $-$ $\gamma $ = 3
C.
$\delta $ + $\beta $ = 4
D.
$\alpha $ + $\beta $ + $\gamma $ = $\delta $
2020
JEE Advanced
MSQ
JEE Advanced 2020 Paper 1 Offline
Let L1 and L2 be the following straight lines.
${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$ and ${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$.
Suppose the straight line
$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$
lies in the plane containing L1 and L2 and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$ and ${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$.
Suppose the straight line
$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$
lies in the plane containing L1 and L2 and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
A.
$\alpha $ $-$ $\gamma $ = 3
B.
l + m = 2
C.
$\alpha $ $-$ $\gamma $ = 1
D.
l + m = 0
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines
$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$ and $\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$ is :
$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$ and $\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$ is :
A.
${1 \over 3}$
B.
${1 \over {\sqrt 3 }}$
C.
3
D.
${\sqrt 3 }$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0,
passes through the point :
A.
(1, –4, 1)
B.
(1, 4, –1)
C.
(2, 4, 1)
D.
(2, –4, 1)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
If the line ${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}}$
intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane
3x + y + 4z = 16 at a point Q, then PQ is equal to :
A.
$2\sqrt 7 $
B.
14
C.
$2\sqrt {14} $
D.
$\sqrt {14} $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
A perpendicular is drawn from a point on the line ${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}$ to the plane x + y + z = 3 such that the
foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :
A.
(4, 0, – 1)
B.
(2, 0, 1)
C.
(1, 0, 2)
D.
(– 1, 0, 4)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
If the plane 2x – y + 2z + 3 = 0 has the distances
${1 \over 3}$
and
${2 \over 3}$
units from the planes 4x – 2y + 4z + $\lambda $ = 0 and
2x – y + 2z + $\mu $ = 0, respectively, then the maximum value of $\lambda $ + $\mu $ is equal to :
A.
13
B.
9
C.
5
D.
15
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
If the length of the perpendicular from the point ($\beta $, 0, $\beta $) ($\beta $ $ \ne $ 0) to the line,
${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$ is $\sqrt {{3 \over 2}} $, then $\beta $ is equal to :
${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$ is $\sqrt {{3 \over 2}} $, then $\beta $ is equal to :
A.
2
B.
1
C.
-2
D.
-1
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
If Q(0, –1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the
area (in sq. units) of $\Delta $PQR is :
A.
${{\sqrt {65} } \over 2}$
B.
$2\sqrt {13} $
C.
${{\sqrt {91} } \over 2}$
D.
${{\sqrt {91} } \over 4}$