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JEE • MATHEMATICS • DCS AND DRS

Dcs and drs practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-45 of 45 questions
31 Easy
31 of 45
Understanding +4 | -1
The distance between two points PP and QQ is dd and the length of their projections of PQP Q on the co-ordinate planes are d1,d2,d3d_1, d_2, d_3. Then d12+d22+d32=kd2d_1^2+d_2^2+d_3^2=k d^2 where ' kk ' is
View answer and explanation
Correct option: D
2
Option D (2) is correct because the sum of the squares of the projections is equal to 2 times the square of the distance between the points.
32 Easy
32 of 45
Analyzing +4 | -1
The angle between the lines whose direction cosines are given by 2l−m+2n=0,lm+mn+nl=02 l-m+2 n=0, l m+m n+n l=0 is
View answer and explanation
Correct option: D
π2\frac{\pi}{2}
Option D (π2\frac{\pi}{2}) is correct because the direction ratios of the lines are perpendicular, resulting in an angle of 90 degrees.
33 Easy
33 of 45
Application +4 | -1
If a line makes angle α,β\alpha, \beta and γ\gamma with the coordinates axes, then
View answer and explanation
Correct option: C
cos⁡2α+cos⁡2β+cos⁡2γ+1=0\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma+1=0
Option C (cos⁡2α+cos⁡2β+cos⁡2γ+1=0\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma+1=0) is correct because it directly uses the identity relating direction cosines and trigonometric identities.
34 Easy
34 of 45
Analyzing +4 | -1
Consider the following 3 lines in space
L1:r=3i−j+2k+λ(2i+4j−k){L_1}:r = 3i - j + 2k + \lambda (2i + 4j - k)
L2:r=i+j−3k+μ(4i+2j+4k){L_2}:r = i + j - 3k + \mu (4i + 2j + 4k)
L3:r=3i+2j−2k+t(2i+j+2k){L_3}:r = 3i + 2j - 2k + t(2i + j + 2k)
Then, which one of the following pair(s) is/are in the same plane?
View answer and explanation
Correct option: D
L1L2L_1 L_2 and L2L3L_2 L_3
Option D (L1L2L_1 L_2 and L2L3L_2 L_3) is correct because L1L_1 and L2L_2 intersect, and L2L_2 and L3L_3 are parallel, implying they are coplanar.
35 Easy
35 of 45
Application +4 | -1
If direction cosines of two lines are proportional to (2,3,−6)(2,3,-6) and (3,−4,5)(3,-4,5), then the acute angle between them is
View answer and explanation
Correct option: B
cos⁡−1(18235)\cos ^{-1}\left(\frac{18 \sqrt{2}}{35}\right)
Option B (cos⁡−1(18235)\cos ^{-1}\left(\frac{18 \sqrt{2}}{35}\right)) is correct because it is calculated using the formula for the angle between two lines with given direction ratios.
36 Easy
36 of 45
Application +4 | -1
If the direction cosines of a line are (1c,1c,1c)\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right) then
View answer and explanation
Correct option: D
c=±3c= \pm \sqrt{3}
Option D (c=±3c= \pm \sqrt{3}) is correct because the sum of the squares of the direction cosines must equal 1.
37 Easy
37 of 45
Application +4 | -1
If the co-ordinates of the points PP and QQ be (1, 2, 1) and (2,3,4)(2,3,4) and OO be the origin, then
View answer and explanation
Correct option: B
OP⊥OQO P \perp O Q
Option B (OP⊥OQO P \perp O Q) is correct because the dot product of the vectors OP and OQ is zero, indicating they are perpendicular.
38 Easy
38 of 45
Application +4 | -1
The plane containing the two lines x−31=y−24=z−15\frac{x-3}{1}=\frac{y-2}{4}=\frac{z-1}{5} and x−21=y+3−4=z+15\frac{x-2}{1}=\frac{y+3}{-4}=\frac{z+1}{5} is 11x+my+nz=2811 x+m y+n z=28 where-
View answer and explanation
Correct option: C
m=−1,n=−3m=-1, n=-3
Option C (m=−1,n=−3m=-1, n=-3) is correct because substituting the points (3, 2, 1) and (2, -3, -1) into the plane equation yields m = -1 and n = -3.
39 Easy
39 of 45
Analyzing +4 | -1
The vector equations of two lines L1L_1 and L2L_2 are respectively
r=17i−9j+9k+λ(3i+j+5k)r = 17i - 9j + 9k + \lambda (3i + j + 5k)
and r=15i−8j−k+μ(4i+3j)r = 15i - 8j - k + \mu (4i + 3j)
I. L1L_1 and L2L_2 are skew lines.

II. (11,−11,−1)(11,-11,-1) is the point of intersection of L1L_1 and L2L_2.

III. (−11,11,1)(-11,11,1) is the point of intersection of L1L_1 and L2L_2.

IV. cos⁡−1(335)\cos ^{-1}\left(\frac{3}{\sqrt{35}}\right) is the acute angle between, L1L_1 and L2L_2.

Then, which of the following is true?
View answer and explanation
Correct option: B
I and IV
Option B (I and IV) is correct because lines L1 and L2 are skew lines, and the acute angle between them is calculated as cos-1(3/sqrt(35)).
40 Easy
40 of 45
Application +4 | -1
If the distance between line 3(x+3)=2(y−5)=z+13(x+3)=2(y-5)=z+1 and a line passing through (1,−2,0)(1,-2,0) and having direction ratios (2,3,6)(2,3,6) is KK units, then K2K^2 is equal to
View answer and explanation
Correct option: A
65
Option A (65) is correct because it is the square of the distance between the two lines, calculated using the formula for the distance between skew lines.
41 Easy
41 of 45
Application +4 | -1
The projection of a line PQP Q, where PP is (0,−1,3)(0,-1,3) and QQ is (4,5,6)(4,5,6) on a line whose direction ratios are 1,2,31,2,3 is
View answer and explanation
Correct option: C
2514\frac{25}{\sqrt{14}}
Option C (2514\frac{25}{\sqrt{14}}) is correct because it represents the projection of the line PQ onto the given direction ratios, calculated using the formula for projection.
42 Easy
42 of 45
Application +4 | -1
The projection of a line segment on the coordinate axes are 2, 3, 6. Then, the length of the line segment is
View answer and explanation
Correct option: A
7
Option A (7) is correct because the length of the line segment is the square root of the sum of the squares of its projections on the coordinate axes.
43 Easy
43 of 45
Application +4 | -1
If the direction cosines of a line are (1c,1c,1c)\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right) then
View answer and explanation
Correct option: D
c=±3c= \pm \sqrt{3}
Option D (c=±3c= \pm \sqrt{3}) is correct because the sum of the squares of the direction cosines must equal 1.
44 Easy
44 of 45
Application +4 | -1
The projections of a vector on the three coordinate axes are 6,−3,26,-3,2 respectively. The direction cosines of the vector are
View answer and explanation
Correct option: C
67,−37,27\frac{6}{7},-\frac{3}{7}, \frac{2}{7}
Option C (67,−37,27\frac{6}{7},-\frac{3}{7}, \frac{2}{7}) is correct because these are the direction cosines of the vector, calculated by dividing the projections by the magnitude of the vector.
45 Easy
45 of 45
Application +4 | -1
The symmetric form of the equations of the line x+y−z=1,2x−3y+z=2x+y-z=1,2 x-3 y+z=2 is
View answer and explanation
Correct option: C
x−12=y3=z5\frac{x-1}{2}=\frac{y}{3}=\frac{z}{5}
Option C (x−12=y3=z5\frac{{x-1}}{{2}}=\frac{{y}}{{3}}=\frac{{z}}{{5}}) is correct because it represents the symmetric form of the line obtained by finding the direction ratios and a point on the line.
Revision notes and key takeaways

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