If a line makes an angle of $30^\circ$ with the positive direction of x-axis, $120^\circ$ with the positive direction of y-axis, then the angle which it makes with the positive direction of z-axis is: (A) $90^\circ$ (B) $120^\circ$ (C) $60^\circ$ (D) $0^\circ$

If a line makes an angle of 30^° with the positive direction of x-axis, 120^° with the positive direction of y

Question

30^\circ

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

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The sum of the squares of the direction cosines of a line is always 1. Using this property, we find that the angle the line makes with the z-axis is $\boxed{90^\circ}$ .

Concept and Intuition

In three-dimensional space, the orientation of a line can be uniquely described by the angles it makes with the positive directions of the x, y, and z axes. Let these angles be $\alpha$ , $\beta$ , and $\gamma$ respectively.

The cosines of these angles, $\cos\alpha$ , $\cos\beta$ , and $\cos\gamma$ , are called the direction cosines of the line. They are usually denoted by $l, m, n$ :

$l = \cos\alpha$

$m = \cos\beta$

$n = \cos\gamma$

These direction cosines are fundamental because they are essentially the components of a unit vector that points in the same direction as the line. Imagine a vector $\vec{v}$ along the line, starting from the origin. If its components are $(x, y, z)$ , then its magnitude is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$ .

The cosine of the angle $\alpha$ that $\vec{v}$ makes with the x-axis is given by $\cos\alpha = \frac{x}{|\vec{v}|}$ . Similarly, $\cos\beta = \frac{y}{|\vec{v}|}$ and $\cos\gamma = \frac{z}{|\vec{v}|}$ .

l^2 + m^2 + n^2 = 1

This property allows us to find an unknown angle if the other two are known, as in this problem.

Step-by-Step Solution

Identify the given angles:

We are given the angles the line makes with the positive x-axis and y-axis:

$\alpha = 30^\circ$

$\beta = 120^\circ$

We need to find $\gamma$ , the angle it makes with the positive z-axis.
Calculate the direction cosines $l$ and $m$ :

Using the definitions of direction cosines:

$l = \cos\alpha = \cos 30^\circ = \frac{\sqrt{3}}{2}$

$m = \cos\beta = \cos 120^\circ$

To evaluate $\cos 120^\circ$ , we can use the identity $\cos(180^\circ - \theta) = -\cos\theta$ :

$\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$

So, $m = -\frac{1}{2}$ .
Apply the fundamental identity for direction cosines:

The sum of the squares of the direction cosines is 1:

$l^2 + m^2 + n^2 = 1$
Substitute the known values and solve for $n^2$ :

Substitute $l = \frac{\sqrt{3}}{2}$ and $m = -\frac{1}{2}$ into the identity: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 84 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$\frac{2x-1}{4} = \frac{1-y}{3} = \frac{-z}{6}$
›Reveal solutionSolution
To find direction cosines, first convert the line's equation to the standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ . The denominators $(a, b, c)$ are the direction ratios. Normalize these ratios by dividing by their magnitude $\sqrt{a^2+b^2+c^2}$ to get the direction cosines. The direction cosines are $\boxed{\frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}}$ .

Concept and Intuition

A line in 3D space has a specific orientation, which can be described by its direction. This direction is represented by a vector parallel to the line.

Direction Ratios: If a vector $\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}$ is parallel to a line, then the numbers $(a, b, c)$ are called the direction ratios of the line. There are infinitely many sets of direction ratios for a given line (e.g., $(2a, 2b, 2c)$ would also be direction ratios).

Direction Cosines: These are a unique set of direction ratios that are normalized. If $(a, b, c)$ are direction ratios, then the direction cosines $(l, m, n)$ are given by:

$l = \frac{a}{\sqrt{a^2+b^2+c^2}}$

$m = \frac{b}{\sqrt{a^2+b^2+c^2}}$

$n = \frac{c}{\sqrt{a^2+b^2+c^2}}$

The direction cosines are essentially the components of a unit vector parallel to the line. They are the cosines of the angles the line makes with the positive $x, y, z$ axes, respectively. An important property is that $l^2 + m^2 + n^2 = 1$ .

The standard symmetric form of the equation of a line passing through a point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$ is:
$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$
The key insight here is that for the denominators to represent the direction ratios, the numerators must be in the form $(x-x_1)$ , $(y-y_1)$ , and $(z-z_1)$ . If they are not, we must algebraically manipulate the equation to achieve this form first.

Step-by-Step Solution

Convert the given equation to standard symmetric form.

The given equation is $\frac{2x-1}{4} = \frac{1-y}{3} = \frac{-z}{6}$ .

We need to transform each part so that the numerators are of the form $(x-x_1)$ , $(y-y_1)$ , and $(z-z_1)$ .

For the first part, $\frac{2x-1}{4}$ :

Factor out 2 from the numerator: $\frac{2(x - 1/2)}{4}$ .

Simplify: $\frac{x - 1/2}{2}$ .

For the second part, $\frac{1-y}{3}$ :

Factor out -1 from the numerator: $\frac{-(y - 1)}{3}$ .

Move the negative sign to the denominator: $\frac{y - 1}{-3}$ .

For the third part, $\frac{-z}{6}$ :

Factor out -1 from the numerator: $\frac{-(z - 0)}{6}$ .

Move the negative sign to the denominator: $\frac{z - 0}{-6}$ .

Now, the equation in standard symmetric form is:

$\frac{x - 1/2}{2} = \frac{y - 1}{-3} = \frac{z - 0}{-6}$ …
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CBSE 2026Set 65/2/11 markMCQ
$< 1, 1, 1 >$
›Reveal solutionSolution
A line’s direction cosines must satisfy $l^2 + m^2 + n^2 = 1$ . Since $1^2+1^2+1^2 = 3 \neq 1$ , the triple $<1,1,1>$ cannot be direction cosines. So Assertion (A) is false. Reason (R) is true because $\cos 0 = 1$ , but it does not explain (A). The correct option is (D).

The core idea here is the definition of direction cosines. Direction cosines of a line are the cosines of the angles the line makes with the coordinate axes. If a line makes angles $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively, then its direction cosines are $l = \cos\alpha$ , $m = \cos\beta$ , $n = \cos\gamma$ .

A fundamental property — and the one that decides this question — is that these three numbers always satisfy $l^2 + m^2 + n^2 = 1$ . Why? Because the direction vector of the line has components proportional to $l, m, n$ , and its magnitude squared equals $l^2 + m^2 + n^2$ times some scale factor; but since $l, m, n$ are themselves the cosines, the vector $(\cos\alpha, \cos\beta, \cos\gamma)$ is a unit vector. So the sum of squares must be exactly 1.

Now let’s examine the Assertion and Reason separately.

Check Assertion (A): Can $<1, 1, 1>$ be direction cosines?

Compute $1^2 + 1^2 + 1^2 = 3$ . This is not equal to 1. Therefore $<1,1,1>$ violates the necessary condition. So the Assertion is false.

Check Reason (R): Is $\cos\theta = 1$ possible?

Yes, $\cos 0 = 1$ . So the statement “ $\cos\theta = 1$ is possible for $\theta = 0$ ” is true.

Does Reason (R) explain Assertion (A)? …
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CBSE 2026Set CX1 mark
$90^{\circ}$
›Reveal solutionSolution

The direction cosines are just the cosines of the given angles: $\left(0,\tfrac12,\tfrac{\sqrt3}{2}\right)$ .

Concept: If a line makes angles $\alpha,\beta,\gamma$ with the $x,y,z$ -axes, its direction cosines are $l=\cos\alpha,\ m=\cos\beta,\ n=\cos\gamma$ .

$l=\cos90^{\circ}=0,\qquad m=\cos60^{\circ}=\dfrac{1}{2},\qquad n=\cos30^{\circ}=\dfrac{\sqrt3}{2}.$

…

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CBSE 2026Set A1 markMCQ
$2, 6, -3$
(a) $\frac{1}{7}, \frac{2}{7}, \frac{3}{7}$
(b) $\frac{2}{7}, \frac{-6}{7}, \frac{3}{7}$
(c) $\frac{2}{7}, \frac{6}{7}, \frac{-3}{7}$
(d) none of these
›Reveal solutionSolution

Direction cosines $=$ direction ratios divided by their magnitude.

Direction ratios are $2, 6, -3$ . Their magnitude is

$\sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7.$ …

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CBSE 2026Set A1 markMCQ
$\alpha$
(a) $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
(b) $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 4$
(c) $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 2$
(d) $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 1$
›Reveal solutionSolution

For direction cosines, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ .

If a line makes angles $\alpha,\beta,\gamma$ with the axes, then $l=\cos\alpha$ , $m=\cos\beta$ , $n=\cos\gamma$ are its direction cosines and satisfy $l^2+m^2+n^2 = 1$ , i.e.

$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.$ …

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CBSE 2026Set ANNUAL1 markMCQ
$30^{\circ}$
(a) $45^{\circ}$
(b) $60^{\circ}$
(c) $120^{\circ}$
(d) Cannot be determined
›Reveal solutionSolution

Applying the direction-cosine identity to the given angles gives a negative value for $\cos^2\gamma$ , which is impossible — so the required angle cannot exist / be determined from the given data.

For a line making angles $\alpha,\beta,\gamma$ with the $X$ -, $Y$ -, $Z$ -axes respectively, the direction cosines $l=\cos\alpha,\ m=\cos\beta,\ n=\cos\gamma$ must satisfy

$l^2+m^2+n^2=1$

Given $\alpha=30^{\circ}$ , $\beta=45^{\circ}$ :

$\cos^2 30^{\circ}=\left(\frac{\sqrt3}{2}\right)^2=\frac34,\qquad \cos^2 45^{\circ}=\left(\frac{1}{\sqrt2}\right)^2=\frac12$

So

$n^2=\cos^2\gamma=1-\frac34-\frac12=1-\frac54=-\frac14$

…

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CBSE 2026Set ANNUAL1 mark
$(-2, 4, -5)$
›Reveal solutionSolution

Find direction ratios from the two points, then divide by their magnitude to get direction cosines.

Direction ratios: $(1-(-2),\,2-4,\,3-(-5)) = (3,-2,8)$ .

Magnitude $=\sqrt{3^2+(-2)^2+8^2}=\sqrt{9+4+64}=\sqrt{77}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\alpha, \beta, \gamma$
(a) 2
(b) 1
(c) -2
(d) 0
›Reveal solutionSolution

The direction cosines of a line satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ ; convert to sines using $\sin^2\theta=1-\cos^2\theta$ .

Since $\alpha,\beta,\gamma$ are the angles a line makes with the coordinate axes, its direction cosines satisfy:

$\cos^2\alpha+\cos^2\beta+\cos^2\gamma = 1$ …

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CBSE 2026Set ANNUAL1 mark
$y$
›Reveal solutionSolution

The $y$ -axis makes angles $90°, 0°, 90°$ with the $x$ , $y$ , $z$ axes respectively.

The direction cosines of a line are $(\cos\alpha,\cos\beta,\cos\gamma)$ , the cosines of the angles it makes with the positive $x$ , $y$ , $z$ axes.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\pm\left(\dfrac{1}{\sqrt3}, \dfrac{1}{\sqrt3}, \dfrac{1}{\sqrt3}\right)$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution

Equal angles with the axes plus the identity $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ together give the direction cosines in Assertion, using the definition in Reason.

Reason (R): If a vector makes angles $\alpha,\beta,\gamma$ with the positive $x$ , $y$ , $z$ axes, its direction cosines are $\cos\alpha,\cos\beta,\cos\gamma$ — this is the standard definition, so R is true.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Direction cosines of x-axis are$
(a) <0, 0, 0>
(b) <1, 1, 1>
(c) <0, 0, 1>
(d) <1, 0, 0>
›Reveal solutionSolution

The x-axis makes a $0°$ angle with itself and $90°$ with both the y- and z-axes.

Direction cosines are $(\cos\alpha,\cos\beta,\cos\gamma)$ where $\alpha,\beta,\gamma$ are the angles the line makes with the x-, y-, z-axes respectively.

For the x-axis itself: $\alpha=0°,\ \beta=90°,\ \gamma=90°$ , so …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): If a line has direction ratios -18, 12, -4 then its direction cosines are -9/11, 6/11, -2/11. Reason (R): If a line has direction ratios a, b, c then its direction cosines are a/\sqrt(a²+b²+c²), b/\sqrt(a²+b²+c²), c/\sqrt(a²+b²+c²).$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Both Assertion (A) and Reason (R) are false.
›Reveal solutionSolution

Dividing the given direction ratios by their magnitude reproduces exactly the direction cosines stated in the Assertion, confirming both statements and the explanation link.

Reason (R) states the standard formula: if direction ratios are $a,b,c$ , the direction cosines are $\dfrac{a}{\sqrt{a^2+b^2+c^2}},\dfrac{b}{\sqrt{a^2+b^2+c^2}},\dfrac{c}{\sqrt{a^2+b^2+c^2}}$ — this is the correct general formula, so R is true.

Checking Assertion (A): ratios are $a=-18,\ b=12,\ c=-4$ .

$\sqrt{a^2+b^2+c^2} = \sqrt{324+144+16} = \sqrt{484} = 22.$ …

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Q.If a line makes an angle of $30^\circ$ with the positive direction of x-axis, $120^\circ$ with the positive direction of y-axis, then the angle which it makes with the positive direction of z-axis is: (A) $90^\circ$ (B) $120^\circ$ (C) $60^\circ$ (D) $0^\circ$

Concept understanding — Direction Cosines Properties

Direction Cosines and Their Properties

Property 1 — the squares sum to 1

Property 2 — they are a unit vector

Property 3 — fixed up to sign

Direction ratios

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

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