iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 26th July Morning Shift
Let $\mathrm{Q}$ and $\mathrm{R}$ be two points on the line $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$ at a distance $\sqrt{26}$ from the point $P(4,2,7)$. Then the square of the area of the triangle $P Q R$ is ___________.
$\therefore$ Area of $\Delta PQR = {1 \over 2} \times 2\sqrt {17} \times 3 = 3\sqrt {17} $
$\therefore$ Square of $ar(\Delta PQR) = 153$.
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 25th July Morning Shift
The line of shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of $\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$ with the plane $\mathrm{P}: \mathrm{a} x-y-z=0$, $(a>0)$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta-\gamma$ is equal to _________________.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 30th June Morning Shift
Consider a triangle ABC whose vertices are A(0, $\alpha$, $\alpha$), B($\alpha$, 0, $\alpha$) and C($\alpha$, $\alpha$, 0), $\alpha$ > 0. Let D be a point moving on the line x + z $-$ 3 = 0 = y and G be the centroid of $\Delta$ABC. If the minimum length of GD is $\sqrt {{{57} \over 2}} $, then $\alpha$ is equal to ____________.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 29th June Morning Shift
Let d be the distance between the foot of perpendiculars of the points P(1, 2, $-$1) and Q(2, $-$1, 3) on the plane $-$x + y + z = 1. Then d2 is equal to ___________.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 29th June Morning Shift
Let ${P_1}:\overrightarrow r \,.\,\left( {2\widehat i + \widehat j - 3\widehat k} \right) = 4$ be a plane. Let P2 be another plane which passes through the points (2, $-$3, 2), (2, $-$2, $-$3) and (1, $-$4, 2). If the direction ratios of the line of intersection of P1 and P2 be 16, $\alpha$, $\beta$, then the value of $\alpha$ + $\beta$ is equal to ________________.
Correct Answer: 28
Explanation:
Direction ratio of normal to ${P_1} \equiv < 2,1, - 3 > $
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th June Evening Shift
Let the image of the point P(1, 2, 3) in the line $L:{{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3}$ be Q. Let R ($\alpha$, $\beta$, $\gamma$) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 ($\alpha$ + $\beta$ + $\gamma$) is equal to __________.
Correct Answer: 125
Explanation:
The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 27th June Morning Shift
Let the mirror image of the point (a, b, c) with respect to the plane 3x $-$ 4y + 12z + 19 = 0 be (a $-$ 6, $\beta$, $\gamma$). If a + b + c = 5, then 7$\beta$ $-$ 9$\gamma$ is equal to ______________.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 25th June Evening Shift
Let l1 be the line in xy-plane with x and y intercepts ${1 \over 8}$ and ${1 \over {4\sqrt 2 }}$ respectively, and l2 be the line in zx-plane with x and z intercepts $ - {1 \over 8}$ and $ - {1 \over {6\sqrt 3 }}$ respectively. If d is the shortest distance between the line l1 and l2, then d$-$2 is equal to _______________.
$d = \left| {{{(\overrightarrow c - \overrightarrow a )\,.\,(\overrightarrow b \times \overrightarrow d )} \over {\left| {\overrightarrow b \times \overrightarrow d } \right|}}} \right|$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 25th June Morning Shift
Let the lines
${L_1}:\overrightarrow r = \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right),\,\lambda \in R$
${L_2}:\overrightarrow r = \left( {\widehat i + 3\widehat j + \widehat k} \right) + \mu \left( {\widehat i + \widehat j + 5\widehat k} \right);\,\mu \in R$,
intersect at the point S. If a plane ax + by $-$ z + d = 0 passes through S and is parallel to both the lines L1 and L2, then the value of a + b + d is equal to ____________.
Correct Answer: 5
Explanation:
As plane is parallel to both the lines we have d.r's of normal to the plane as <7, $-$2, $-$1>
Also point of intersection of lines is $2\widehat i + 4\widehat j + 6\widehat k$
$\therefore$ Equation of plane is
$7(x - 2) - 2(y - 4) - 1(z - 6) = 0$
$ \Rightarrow 7x - 2y - z = 0$
$a + b + d = 7 - 2 + 0 = 5$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 24th June Morning Shift
Let a line having direction ratios, 1, $-$4, 2 intersect the lines ${{x - 7} \over 3} = {{y - 1} \over { - 1}} = {{z + 2} \over 1}$ and ${x \over 2} = {{y - 7} \over 3} = {z \over 1}$ at the points A and B. Then (AB)2 is equal to ___________.
Correct Answer: 84
Explanation:
Let $A(3 \lambda+7,-\lambda+1, \lambda-2)$ and $B(2 \mu, 3 \mu+7, \mu)$
So, DR's of $A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu$ $-2$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 24th June Morning Shift
If the shortest distance between the lines
$\overrightarrow r = \left( { - \widehat i + 3\widehat k} \right) + \lambda \left( {\widehat i - a\widehat j} \right)$
and $\overrightarrow r = \left( { - \widehat j + 2\widehat k} \right) + \mu \left( {\widehat i - \widehat j + \widehat k} \right)$ is $\sqrt {{2 \over 3}} $, then the integral value of a is equal to ___________.
$
\begin{aligned}
&\Rightarrow \quad \sqrt{\frac{2}{3}}=\frac{2(a-1)}{\sqrt{a^{2}+1+(a-1)^{2}}} \\\\
&\Rightarrow 6\left(a^{2}-2 a+1\right)=2 a^{2}-2 a+2 \\\\
&\Rightarrow \quad(a-2)(2 a-1)=0 \Rightarrow a=2 \text { because } a \in z .
\end{aligned}
$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 1st September Evening Shift
Let the acute angle bisector of the two planes x $-$ 2y $-$ 2z + 1 = 0 and 2x $-$ 3y $-$ 6z + 1 = 0 be the plane P. Then which of the following points lies on P?
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 31st August Morning Shift
Let the equation of the plane, that passes through the point (1, 4, $-$3) and contains the line of intersection of the planes 3x $-$ 2y + 4z $-$ 7 = 0 and x + 5y $-$ 2z + 9 = 0, be $\alpha$x + $\beta$y + $\gamma$z + 3 = 0, then $\alpha$ + $\beta$ + $\gamma$ is equal to :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th August Evening Shift
The equation of the plane passing through the line of intersection of the planes $\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$ and $\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0$ and parallel to the x-axis is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th August Morning Shift
Equation of a plane at a distance $\sqrt {{2 \over {21}}} $ from the origin, which contains the line of intersection of the planes x $-$ y $-$ z $-$ 1 = 0 and 2x + y $-$ 3z + 4 = 0, is :
A.
$3x - y - 5z + 2 = 0$
B.
$3x - 4z + 3 = 0$
C.
$ - x + 2y + 2z - 3 = 0$
D.
$4x - y - 5z + 2 = 0$
Correct Answer: D
Explanation:
Required equation of plane
${P_1} + \lambda {P_2} = 0$
$(x - y - z - 1) + \lambda (2x + y - 3z + 4) = 0$
Given that its dist. From origin is ${2 \over {\sqrt {21} }}$
for $\lambda = {1 \over 2}$ reqd. plane is $4x - y - 5z + 2 = 0$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Evening Shift
Let P be the plane passing through the point (1, 2, 3) and the line of intersection of the planes $\overrightarrow r \,.\,\left( {\widehat i + \widehat j + 4\widehat k} \right) = 16$ and $\overrightarrow r \,.\,\left( { - \widehat i + \widehat j + \widehat k} \right) = 6$. Then which of the following points does NOT lie on P?
A.
(3, 3, 2)
B.
(6, $-$6, 2)
C.
(4, 2, 2)
D.
($-$8, 8, 6)
Correct Answer: C
Explanation:
$(x + y + 4z - 16) + \lambda ( - x + y + z - 6) = 0$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Morning Shift
A plane P contains the line $x + 2y + 3z + 1 = 0 = x - y - z - 6$, and is perpendicular to the plane $ - 2x + y + z + 8 = 0$. Then which of the following points lies on P?
A.
($-$1, 1, 2)
B.
(0, 1, 1)
C.
(1, 0, 1)
D.
(2, $-$1, 1)
Correct Answer: B
Explanation:
Equation of plane P can be assumed as
P : x + 2y + 3z + 1 + $\lambda$ (x $-$ y $-$ z $-$ 6) = 0
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Evening Shift
For real numbers $\alpha$ and $\beta$ $\ne$ 0, if the point of intersection of the straight lines
${{x - \alpha } \over 1} = {{y - 1} \over 2} = {{z - 1} \over 3}$ and ${{x - 4} \over \beta } = {{y - 6} \over 3} = {{z - 7} \over 3}$, lies on the plane x + 2y $-$ z = 8, then $\alpha$ $-$ $\beta$ is equal to :
A.
5
B.
9
C.
3
D.
7
Correct Answer: D
Explanation:
First line is ($\phi$ + $\alpha$, 2$\phi$ + 1, 3$\phi$ + 1)
and second line is (q$\beta$ + 4, 3q + 6, 3q + 7)
For intersection
$\phi$ + $\alpha$ = q$\beta$ + 4 ...... (i)
2$\phi$ + 1 = 3q + 6 .... (ii)
3$\phi$ + 1 = 3q + 7 ...... (iii)
for (ii) & (iii) $\phi$ = 1, q = $-$1
So, from (i) $\alpha$ + $\beta$ = 3
Now, point of intersection is ($\alpha$ + 1, 3, 4)
It lies on the plane.
Hence, $\alpha$ = 5 & $\beta$ = $-$2
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Morning Shift
Let the plane passing through the point ($-$1, 0, $-$2) and perpendicular to each of the planes 2x + y $-$ z = 2 and x $-$ y $-$ z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to :
A.
3
B.
8
C.
5
D.
4
Correct Answer: D
Explanation:
Normal of required plane $\left( {2\widehat i + \widehat j - \widehat k} \right) \times \left( {\widehat i - \widehat j - \widehat k} \right)$
$ = - 2\widehat i + \widehat j - 3\widehat k$
Equation of plane
$ - 2(x + 1) + 1(y - 0) - 3(z + 2) = 0$
$ - 2x + y - 3z - 8 = 0$
$2x - y + 3z + 8 = 0$
$a + b + c = 4$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th July Morning Shift
Let the foot of perpendicular from a point P(1, 2, $-$1) to the straight line $L:{x \over 1} = {y \over 0} = {z \over { - 1}}$ be N. Let a line be drawn from P parallel to the plane x + y + 2z = 0 which meets L at point Q. If $\alpha$ is the acute angle between the lines PN and PQ, then cos$\alpha$ is equal to ________________.
A.
${1 \over {\sqrt 5 }}$
B.
${{\sqrt 3 } \over 2}$
C.
${1 \over {\sqrt 3 }}$
D.
${1 \over {2\sqrt 3 }}$
Correct Answer: C
Explanation:
$\overrightarrow {PN} .(\widehat i - \widehat k) = 0$
$\overrightarrow {PN} $ = 2$\widehat j$ and $\overrightarrow {PQ} $ = $2\widehat i + 2\widehat j - 2\widehat k$
$ \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 22th July Evening Shift
Let L be the line of intersection of planes $\overrightarrow r .(\widehat i - \widehat j + 2\widehat k) = 2$ and $\overrightarrow r .(2\widehat i + \widehat j - \widehat k) = 2$. If $P(\alpha ,\beta ,\gamma )$ is the foot of perpendicular on L from the point (1, 2, 0), then the value of $35(\alpha + \beta + \gamma )$ is equal to :
A.
101
B.
119
C.
143
D.
134
Correct Answer: B
Explanation:
${P_1}:x - y + 2z = 2$
${P_2}:2x + y - 3 = 2$
Let line of Intersection of planes P1 and P2 cuts xy plane in point Q.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 22th July Evening Shift
If the shortest distance between the straight lines $3(x - 1) = 6(y - 2) = 2(z - 1)$ and $4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R$ is ${1 \over {\sqrt {38} }}$, then the integral value of $\lambda$ is equal to :
Let Q be the mirror image of the point (2, 3, $-$1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?
A.
($-$1, 1, 2)
B.
(1, 1, 1)
C.
(1, 1, 2)
D.
(1, 2, 2)
Correct Answer: D
Explanation:
Plane p is ${ \bot ^r}$ to line ${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$ & passes through pt. (2, 3) equation of plane p
2(x $-$ 2) + 1(y $-$ 3) + 1 (z + 1) = 0
2x + y + z $-$ 6 = 0
Point (1, 2, 2) satisfies above equation
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 17th March Evening Shift
If the equation of plane passing through the mirror image of a point (2, 3, 1) with respect to line ${{x + 1} \over 2} = {{y - 3} \over 1} = {{z + 2} \over { - 1}}$ and containing the line ${{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1}$ is $\alpha$x + $\beta$y + $\gamma$z = 24, then $\alpha$ + $\beta$ + $\gamma$ is equal to :
A.
21
B.
19
C.
18
D.
20
Correct Answer: B
Explanation:
Let point M is (2$\lambda$ $-$ 1, $\lambda$ + 3, $-$ $\lambda$ $-$ 2)
D.R.'s of AM line are < 2$\lambda$ $-$ 1 $-$ 2, $\lambda$ + 3 $-$ 3, $-$$\lambda$ $-$ 2 $-$ 1>
$ \therefore $ Equation of plane is a(x $-$ 1) + c(z $-$ 3) = 0
x = 0, z = 0 also satisfy it $-$a $-$3c = 0 $ \Rightarrow $ a = $-$3c
$-$3c (x $-$ 1) + c (z $-$ 3) = 0
$-$3 + 3 + z $-$ 3 = 0
3x $-$ z = 0
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Evening Shift
If the foot of the perpendicular from point (4, 3, 8) on the line ${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$, l $\ne$ 0 is (3, 5, 7), then the shortest distance between the line L1 and line ${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$ is equal to :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Evening Shift
If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression $3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {{x + y + z} \over {14(x - 11)(y - 19)(z - 12)}}$ is equal to :
A.
3
B.
39
C.
$-$45
D.
0
Correct Answer: A
Explanation:
From intercept from, equation of plane is x + y + z = 42
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Morning Shift
Let the position vectors of two points P and Q be 3$\widehat i$ $-$ $\widehat j$ + 2$\widehat k$ and $\widehat i$ + 2$\widehat j$ $-$ 4$\widehat k$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, $-$1, 2) and ($-$2, 1, $-$2), respectively. Let lines PR and QS intersect at T. If the vector $\overrightarrow {TA} $ is perpendicular to both $\overrightarrow {PR} $ and $\overrightarrow {QS} $ and the length of vector $\overrightarrow {TA} $ is $\sqrt 5 $ units, then the modulus of a position vector of A is :
A.
$\sqrt {171} $
B.
$\sqrt {227} $
C.
$\sqrt {482} $
D.
$\sqrt {5} $
Correct Answer: A
Explanation:
$\overrightarrow p = 3\widehat i - \widehat j + 2\widehat k$ & $\overrightarrow Q = \widehat i + 2\widehat j - 4\widehat k$
${\overrightarrow v _{PR}} = (4, - 1,2)$ & ${\overrightarrow v _{QS}} = ( - 2,1, - 2)$
${L_{PR}}:\overrightarrow r = (3\widehat i - \widehat j + 2\widehat k) + \lambda (4, - 1,2)$
${L_{QS}}:\overrightarrow r = (\widehat i + 2\widehat j - 4\widehat k) + \mu ( - 2,1, - 2)$
Now T on PR = $\left\langle {3 + 4\lambda , - 1 - \lambda ,2 + 2\lambda } \right\rangle $
Similarly T on QS = (1 $-$ 2$\mu$, 2 + $\mu$, $-$4 $-$ 2$\mu$)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Morning Shift
If for a > 0, the feet of perpendiculars from the points A(a, $-$2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, $-$a, $-$1) and D respectively, then the length of line segment CD is equal to :
A.
$\sqrt {41} $
B.
$\sqrt {55} $
C.
$\sqrt {31} $
D.
$\sqrt {66} $
Correct Answer: D
Explanation:
Let $\phi $ is the angle between $\overrightarrow {AB} $ and $\overrightarrow n $.
CD = AR = | AB |sin$\phi$
CD = | AB | $\sqrt {1 - {{\cos }^2}\phi } $
CD = | AB | $\sqrt {1 - {{\left( {{{\overrightarrow {AB} .\,\overrightarrow n } \over {|AB|}}} \right)}^2}} $
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th February Evening Shift
Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P($\alpha$, $\beta$, $\gamma$) is the foot of perpendicular from (3, 2, 1) on L, then the value of 21($\alpha$ + $\beta$ + $\gamma$) equals :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th February Evening Shift
A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, $-$1, 1), then the projection of $\overrightarrow {OP} $ on this plane is of length :
Any point on this line $(2\lambda + 1,3\lambda - 1, - 2\lambda + 1)$
Direction ratio of given line $(2,3, - 2)$
Direction ratio of line to be found $(2\lambda + 1,3\lambda - 2, - 2\lambda - 1)$
$ \therefore $ ${\overrightarrow d _1}\,.\,{\overrightarrow d _2} = 0$
$ \Rightarrow $ $\lambda = 2/17$
Direction ratio of line $(21, - 28, - 21) \equiv (3, - 4, - 3) \equiv ( - 3,4,3)$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
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.
$\overrightarrow r .\left( {\widehat i - 7\widehat j + 3\widehat k} \right) = {7 \over 3}$
Correct Answer: B
Explanation:
Given, point (1, 0, 2)
Equation of plane =
$\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) = 1$ and $\overrightarrow r\,.\,(\widehat i - 2\widehat j) = - 2$
Equation of plane passing through the intersection of given planes is
Now, the equation of plane passing through (1, 2, -3) having normal
vector -11${\widehat i}$ - ${\widehat j}$ - 17${\widehat k}$ is
-[11(x - 1) + (y - 2) + 17(z + 3)] = 0
$ \Rightarrow $ 11x + y + 17z + 38 = 0
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
iCON Education HYD, 79930 92826, 73309 72826JEE 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.
iCON Education HYD, 79930 92826, 73309 72826JEE 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 ___________.
