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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

The equation of the plane through the intersection of the planes x + y + z = 1 and 2x + 3y $$-$$ z + 4 = 0 and parallel to the x-axis is

A
y + 3z + 6 = 0
B
y + 3z $$-$$ 6 = 0
C
y $$-$$ 3z + 6 = 0
D
y $$-$$ 3z $$-$$ 6 = 0

তলদ্বয় x + y + z = 1 ও 2x + 3y $$-$$ z + 4 = 0 এর ছেদসরলরেখার ধারক ও x-অক্ষের সমান্তরাল তলের সমীকরণ হবে

A
y + 3z + 6 = 0
B
y + 3z $$-$$ 6 = 0
C
y $$-$$ 3z + 6 = 0
D
y $$-$$ 3z $$-$$ 6 = 0
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
The plane lx + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle $$\alpha$$. The equation changes to
A
$$lx + my \pm \tan \alpha \sqrt {{l^2} + {m^2}} = 0$$
B
$$lx + my \pm z\tan \alpha \sqrt {{l^2} + {m^2} + 1} = 0$$
C
$$lx + my \pm z\tan \alpha \sqrt {{l^2} + 1} = 0$$
D
$$lx + my \pm z\tan \alpha \sqrt {{l^2} + {m^2}} = 0$$

Explanation

Given plane,

P1 = lx + my = 0 and P2 = z = 0

Plane through common line of P1 and P2

$$\therefore$$ P3 = lx + my + nz = 0

angle between P1 and P3 = $$\alpha$$

$$\therefore$$ $$\cos \alpha = {{{l^2} + {m^2}} \over {\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{l^2} + {m^2}} }}$$

$$ \Rightarrow \cos \alpha = \sqrt {{{{l^2} + {m^2}} \over {{l^2} + {m^2} + {n^2}}}} $$

$$ \Rightarrow {\cos ^2}\alpha = {{{l^2} + {m^2}} \over {{l^2} + {m^2} + {n^2}}}$$

$$ \Rightarrow {\sec ^2}\alpha = {{{l^2} + {m^2} + {n^2}} \over {{l^2} + {m^2}}} = 1 + {{{n^2}} \over {{l^2} + {m^2}}}$$

$$ \Rightarrow ({\sec ^2}\alpha - 1)({l^2} + {m^2}) = {n^2}$$

$$ \Rightarrow {n^2} = {\tan ^2}\alpha ({l^2} + {m^2}) = n = \pm \sqrt {({l^2} + {m^2})} \tan \alpha $$

$$\therefore$$ Equation becomes

$$lx + my \pm z\tan \alpha \sqrt {{l^2} + {m^2}} = 0$$
$$l$$x + my = 0 তলটি z = 0 তলের সঙ্গে ছেদ সরলরেখা বরাবর $$\alpha$$ কোণে ঘূর্ণন করে। সেক্ষেত্রে সমীকরণটির পরিবর্তিত আকার হবে
A
$$lx + my \pm \tan \alpha \sqrt {{l^2} + {m^2}} = 0$$
B
$$lx + my \pm z\tan \alpha \sqrt {{l^2} + {m^2} + 1} = 0$$
C
$$lx + my \pm z\tan \alpha \sqrt {{l^2} + 1} = 0$$
D
$$lx + my \pm z\tan \alpha \sqrt {{l^2} + {m^2}} = 0$$

Explanation

প্রদত্ত সমতল,

P1 = $$l$$x + my = 0 and P2 = z = 0

P1 এবং P2 এর সাধারণ লাইনের মধ্য দিয়ে সমতল

$$\therefore$$ P3 = lx + my + nz = 0

P1 এবং P3 = $$\alpha$$ এর মধ্যে কোণ

$$\therefore$$ $$\cos \alpha = {{{l^2} + {m^2}} \over {\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{l^2} + {m^2}} }}$$

$$ \Rightarrow \cos \alpha = \sqrt {{{{l^2} + {m^2}} \over {{l^2} + {m^2} + {n^2}}}} $$

$$ \Rightarrow {\cos ^2}\alpha = {{{l^2} + {m^2}} \over {{l^2} + {m^2} + {n^2}}}$$

$$ \Rightarrow {\sec ^2}\alpha = {{{l^2} + {m^2} + {n^2}} \over {{l^2} + {m^2}}} = 1 + {{{n^2}} \over {{l^2} + {m^2}}}$$

$$ \Rightarrow ({\sec ^2}\alpha - 1)({l^2} + {m^2}) = {n^2}$$

$$ \Rightarrow {n^2} = {\tan ^2}\alpha ({l^2} + {m^2}) = n = \pm \sqrt {({l^2} + {m^2})} \tan \alpha $$

$$\therefore$$ সমীকরণ হয়ে যায়

$$lx + my \pm z\tan \alpha \sqrt {{l^2} + {m^2}} = 0$$
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
A line with positive direction cosines passes through the point P(2, $$-$$1, 2) and makes equal angle with co-ordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals.
A
1 unit
B
$$\sqrt 2 $$ unit
C
$$\sqrt 3 $$ unit
D
2 unit

Explanation

Let direction cosines of line is (l, l, l).

$$\therefore$$ l2 + l2 + l2 = 1

$$\Rightarrow$$ 3 l2 = 1 $$\Rightarrow$$ l = $${1 \over {\sqrt 3 }}$$

$$\therefore$$ Direction cosines of line is $$\left( {{1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }}} \right)$$

Now, equation of a line passing through P(2, $$-$$1, 2)

$${{x - 2} \over {1/\sqrt 3 }} = {{y + 1} \over {1/\sqrt 3 }} = {{z - 2} \over {1/\sqrt 3 }} = k$$ .... (i)

$$\Rightarrow$$ x = k + 2, y = k $$-$$ 1 and z = k + 2

Since line (i) meets the plane

2x + y + z = 9 at Q

$$\therefore$$ 2(k + 2) + (k $$-$$ 1) + (k + 2) = 9

$$\Rightarrow$$ 2k + 4 + k $$-$$ 1 + k + 2 = 9

$$\Rightarrow$$ 4k + 5 = 9

$$\Rightarrow$$ 4k = 4 $$\Rightarrow$$ k = 1

$$\therefore$$ Coordinates of Q are (3, 0, 3)

$$\therefore$$ $$PQ = \sqrt {{{(3 - 2)}^2} + {{(0 + 1)}^2} + {{(3 - 2)}^2}} $$

$$ = \sqrt {1 + 1 + 1} = \sqrt 3 $$ units.
ধনাত্মক দিগঙ্কগােষ্ঠী (direction cosines) বিশিষ্ট একটি সরলরেখা P(2,$$-$$1, 2) বিন্দুগামী এবং স্থানাঙ্ক, অক্ষগুলির সঙ্গে সমান কোণ উৎপন্ন করে। ঐ রেখাটি 2x + y + z = 9 তলকে Q বিন্দুতে ছেদ করে। সেক্ষেত্রে PQ রেখাংশের দৈর্ঘ্য হল
A
1 একক
B
$$\sqrt 2 $$ একক
C
$$\sqrt 3 $$ একক
D
2 একক

Explanation

ধরা যাক রেখার দিগঙ্কগােষ্ঠী হল (l, l, l).

$$\therefore$$ l2 + l2 + l2 = 1

$$\Rightarrow$$ 3 l2 = 1 $$\Rightarrow$$ l = $${1 \over {\sqrt 3 }}$$

$$\therefore$$ রেখার দিগঙ্কগােষ্ঠী $$\left( {{1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }}} \right)$$

এখন, P(2, $$−$$1, 2) এর মধ্য দিয়ে যাওয়া একটি রেখার সমীকরণ

$${{x - 2} \over {1/\sqrt 3 }} = {{y + 1} \over {1/\sqrt 3 }} = {{z - 2} \over {1/\sqrt 3 }} = k$$ .... (i)

$$\Rightarrow$$ x = k + 2, y = k $$-$$ 1 and z = k + 2

যেহেতু রেখা (i) সমতলের সাথে মিলিত হয়

Q তে 2x + y + z = 9

$$\therefore$$ 2(k + 2) + (k $$-$$ 1) + (k + 2) = 9

$$\Rightarrow$$ 2k + 4 + k $$-$$ 1 + k + 2 = 9

$$\Rightarrow$$ 4k + 5 = 9

$$\Rightarrow$$ 4k = 4 $$\Rightarrow$$ k = 1

$$\therefore$$ Q-এর স্থানাঙ্ক হল (3, 0, 3)

$$\therefore$$ $$PQ = \sqrt {{{(3 - 2)}^2} + {{(0 + 1)}^2} + {{(3 - 2)}^2}} $$

$$ = \sqrt {1 + 1 + 1} = \sqrt 3 $$ একক.
4

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
If from a point P(a, b, c), perpendicular PA and PB are drawn to YZ and ZX-planes respectively, then the equation of the plane OAB is
A
bcx + cay + abz = 0
B
bcx + cay $$-$$ abz = 0
C
bcx $$-$$ cay + abz = 0
D
bcx $$-$$ cay $$-$$ abz = 0

Explanation



According to question the coordinates of A and B are (0, b, c) and (a, 0, c) respectively.

Now, perpendicular vector OA $$\times$$ OB

$$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & b & c \cr a & 0 & c \cr } } \right|$$

$$ = \widehat i(bc - 0) - \widehat j(0 - ac) + \widehat k(0 - ab)$$

$$ = bc\widehat i + ac\widehat j - ab\widehat k$$

$$\therefore$$ Required equation of plane is bcx + acy $$-$$ abz = 0
P(a, b, c) বিন্দু থেকে YZ-তল ও ZX-তলের উপর লম্ব যথাক্রমে PA ও PB টানা হল। সেক্ষেত্রে OAB তলের সমীকরণ হবে
A
bcx + cay + abz = 0
B
bcx + cay $$-$$ abz = 0
C
bcx $$-$$ cay + abz = 0
D
bcx $$-$$ cay $$-$$ abz = 0

Explanation



প্রশ্ন অনুসারে A এবং B এর স্থানাঙ্ক যথাক্রমে (0, b, c) এবং (a, 0, c)।

এখন, লম্ব ভেক্টর OA $$\times$$ OB

$$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & b & c \cr a & 0 & c \cr } } \right|$$

$$ = \widehat i(bc - 0) - \widehat j(0 - ac) + \widehat k(0 - ab)$$

$$ = bc\widehat i + ac\widehat j - ab\widehat k$$

$$\therefore$$ তলের প্রয়োজনীয় সমীকরণ হল bcx + acy $$-$$ abz = 0

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