If the direction cosines of a straight line are $ $, then

$k=\frac{1}{\sqrt{3}}$ or $-\frac{1}{\sqrt{3}}$

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JEE • MATHEMATICS • DCS AND DRS

Dcs and drs practice · Medium

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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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1 Medium

1 of 16

Application +4 | -1

The reflection of the point

P(1,0,0)

in the line

\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}

View answer and explanation

Correct option: B

(5,-8,-4)

Option B (

(5,-8,-4)

) is correct because it represents the reflection of the point P in the given line, found by determining the foot of the perpendicular and then extending to the reflection point.

2 Medium

2 of 16

Application +4 | -1

Let

{d}

be the distance of the point of intersection of the lines

\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}

and

\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}

from the point

(7,8,9).

Then

d^2+6

is equal to:

View answer and explanation

Correct option: C

The correct answer is C (75) because it's the value of d² + 6, where d is the distance between the intersection point of two lines and the point (7,8,9).

3 Medium

3 of 16

Application +4 | -1

The projection of line joining

(3,4,5)

and

(4,6,3)

on the line joining

(-1,2,4)

and

(1,0,5)

\frac{k}{3}

then '

k

' equals.

View answer and explanation

Correct option: A

Option A (4) is correct because the projection is calculated using the dot product of the direction vectors divided by the magnitude of the line on which the projection is being found, resulting in 4/3, thus k=4.

4 Medium

4 of 16

Application +4 | -1

A line passing through

{A}(1,2,3)

and having direction ratios

(3,4,5)

meets a plane

x+2 y-3 z=5

B

, then distance

A B

is equal to-

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the distance AB calculated using the parametric form of the line and the plane equation does not match any of the given options.

5 Medium

5 of 16

Application +4 | -1

If the direction cosines of a straight line are

< k, k, k >

, then

View answer and explanation

Correct option: D

k=\frac{1}{\sqrt{3}}

-\frac{1}{\sqrt{3}}

Option D (30°) is correct. Since the direction ratios of the line and the normal to the plane are proportional, the line is perpendicular to the plane, making the angle between them 90°. Therefore, the angle between the line and the plane is 90-60 = 30°.

6 Medium

6 of 16

Analyzing +4 | -1

Let

P

be the plane containing the straight line

\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}

and perpendicular to the plane containing the straight lines

\frac{x}{2}=\frac{y}{3}=\frac{z}{5}

and

\frac{x}{3}=\frac{y}{7}=\frac{z}{8}

. If

d

is the distance

P

from the point

(2,-5,11)

, then

d^2

is equal to :

View answer and explanation

Correct option: C

\frac{32}{3}

Option C is correct. The equation of the plane is found using the cross product of the direction vectors and then the distance formula is applied.

7 Medium

7 of 16

Application +4 | -1

Angle between the rays with d.r.'s

4,-3,5 \,\&\, 3, 4,5

is-

View answer and explanation

Correct option: C

\cos ^{-1}\left(\frac{1}{2}\right)

Option C (

\cos ^{-1}(\frac{1}{2})

) is correct because it's the angle between the two rays, calculated using the dot product formula.

8 Medium

8 of 16

Application +4 | -1

If the d.r's of the two perpendicular lines are

(2,-3,5)

and

(\lambda, 1+\lambda, 2+\lambda)

then

\lambda=

View answer and explanation

Correct option: C

-\frac{7}{4}

Option C (-7/4) is correct because for two perpendicular lines, the dot product of their direction ratios is zero. Solving the equation 2λ - 3(1+λ) + 5(2+λ) = 0 gives λ = -7/4.

9 Medium

9 of 16

Analyzing +4 | -1

Let

L_{1}=\frac{x-3}{1}=\frac{y-2}{2}=\frac{z-1}{3}

and

L_{2}=\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}

and direction ratios of line

L_{3}

are

<1,-1,3>. P

and

Q

are points of intersection of

L_{1}

and

L_{3}

and

L_{2} \& L_{3}

, respectively. Then, distance between

P

and

Q

is:

View answer and explanation

Correct option: C

\frac{4}{3} \sqrt{11}

Option C (

\frac{4}{3} \sqrt{11}

) is correct because it represents the distance between points P and Q, calculated using the given line equations and direction ratios.

10 Medium

10 of 16

Application +4 | -1

A vector

\mathbf{r}

is equally inclined with the co-ordinate axes. If the tip of

\mathbf{r}

is in the positive octant and

|\mathbf{r}|=6

, then

\mathbf{r}

View answer and explanation

Correct option: D

2 \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})

Option D,

2 \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})

, is correct because it represents a vector with magnitude 6, equally inclined to the coordinate axes, and lying in the positive octant.

11 Medium

11 of 16

Application +4 | -1

The direction ratios of line

I_1

passing through

P(1,3,4)

and perpendicular to line

I_2 \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}

(where,

I_1

and

I_2

are coplanar) is

View answer and explanation

Correct option: C

14,

-8,-1

Option C (14,

-8,-1

) is correct because it represents the direction ratios of a line passing through P and perpendicular to the given line, considering they are coplanar.

12 Medium

12 of 16

Application +4 | -1

Let

{P}({x}, {y}, {z})

be a point in the first octant, whose projection in the

x y

-plane is the point

{Q}

. Let

{OP}=\gamma

; the angle between

{OQ}

and the positive

x

-axis be

\theta

; and the angle between OP and the positive

{z}

-axis be

\phi

, where

{O}

is the origin. Then the distance of

{P}

from the

{x}

-axis is:

View answer and explanation

Correct option: A

\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}

Option A (

\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}

) is correct because by using the given information and trigonometric relations, we can express the distance of P from the x-axis in terms of

\gamma

\theta

, and

\phi

13 Medium

13 of 16

Application +4 | -1

The angle between the two lines

\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}

\frac{{x}-4}{1}=\frac{{y}+4}{2}=\frac{{z}+1}{2}

View answer and explanation

Correct option: A

\cos ^{-1}\left(\frac{4}{9}\right)

Option A (

\cos ^{-1}\left(\frac{4}{9}\right)

) is correct because it represents the angle between the two given lines, calculated using the direction ratios and the formula for the cosine of the angle between two lines.

14 Medium

14 of 16

Analyzing +4 | -1

The equation of the straight line through the origin and parallel to the line

(b+c) x+(c+a) y

+(a+b) z=k=(b-c) x+(c-a) y+(a-b) z

are

View answer and explanation

Correct option: C

\frac{x}{a^2-b c}=\frac{y}{b^2-c a}=\frac{z}{c^2-a b}

Option C (

\frac{x}{a^2-b c}=\frac{y}{b^2-c a}=\frac{z}{c^2-a b}

) is correct because solving the equations of the planes gives the direction ratios of the line, which match this option.

15 Medium

15 of 16

Application +4 | -1

If a point

R (4, y, z)

lies on the line segment joining the points

P (2, -3, 4)

and

Q (8, 0, 10),

then the distance of

R

from the origin is

View answer and explanation

Correct option: C

2\sqrt {14}

Option C (

2\sqrt {14}

) is correct because after finding the coordinates of point R using the section formula, its distance from the origin is

2\sqrt{14}

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16 Medium

16 of 16

Application +4 | -1

Consider the lines

L_{1}

and

L_{2}

given by

\begin{array}{l} L_{1}: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2} \\ L_{2}: \frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3} \end{array}

A line

L_{3}

having direction ratios,

1,-1,-2

, intersects

L_{1}

and

L_{2}

at the points

P

and

Q

respectively. Then the length of line segment

P Q

View answer and explanation

Correct option: B

2 \sqrt{6}

Option B (

2 \sqrt{{6}}

) is correct because it calculates the distance between the points P and Q where the line L3 intersects L1 and L2, using the given direction ratios.

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Key Takeaways

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