Q.If a line makes angle $90^\circ$, $60^\circ$ and $30^\circ$ with the positive direction of $x$, $y$ and $z$-axis respectively, find its direction cosines.

Showing the 12 most recent of 84 on this concept.

1. CBSE 2024Set 65/1/11 markMCQ
   Q.If the direction cosines of a line are $\sqrt{3}k, \sqrt{3}k, \sqrt{3}k$, then the value of $k$ is : (A) $\pm 1$ (B) $\pm \sqrt{3}$ (C) $\pm 3$ (D) $\pm \frac{1}{3}$

   ›Reveal solutionSolution

   Direction cosines must satisfy $l^2 + m^2 + n^2 = 1$. Substituting $l = m = n = \sqrt{3}k$ gives $3(\sqrt{3}k)^2 = 1 \Rightarrow 9k^2 = 1 \Rightarrow k = \pm \frac{1}{3}$. So the correct option is (D).

   The key idea here is that direction cosines are not just any numbers — they are the cosines of the angles a line makes with the coordinate axes. Because of that, they have a fixed property: the sum of their squares is always exactly 1. This is a non-negotiable condition, and it’s the only tool you need to solve this problem.

   Many students get tempted to treat $\sqrt{3}k$ as a single number and forget to square it properly, or they mistakenly think the sum of the cosines themselves equals 1. That’s a common trap — so let’s be precise.

   1. Recall the fundamental property of direction cosines. If a line has direction cosines $l, m, n$ (with respect to the $x$, $y$, and $z$ axes respectively), then:

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

   This is because the direction cosines are the components of a unit vector along the line.

   2. Substitute the given values. Here, $l = \sqrt{3}k$, $m = \sqrt{3}k$, $n = \sqrt{3}k$. So:

   $(\sqrt{3}k)^2 + (\sqrt{3}k)^2 + (\sqrt{3}k)^2 = 1$

   3. Simplify the squares. $(\sqrt{3}k)^2 = 3k^2$. So the equation becomes:

   $3k^2 + 3k^2 + 3k^2 = 1$

   $9k^2 = 1$

   4. Solve for $k$.

   $k^2 = \frac{1}{9}$

   $k = \pm \frac{1}{3}$

   Watch out

   A common mistake is to write $(\sqrt{3}k)^2 = \sqrt{3}k^2$ or to forget that squaring removes the sign. Always square the entire term, including the coefficient.

   Tip

   Notice that all three direction cosines are equal here. That means the line makes equal angles with all three axes — it’s symmetrically inclined. In such a case, each direction cosine is $\pm \frac{1}{\sqrt{3}}$. Here, $\sqrt{3}k = \pm \frac{1}{\sqrt{3}}$ gives $k = \pm \frac{1}{3}$ directly — a quick mental check.

   ✓Final answer

   The value of $k$ is $\pm \frac{1}{3}$, which corresponds to option (D).

2. CBSE 20201 markMCQ
   Q.If the direction cosines of a line are a, a, a, then (A) a > 0 (B) a = 1 or a = –1 (C) 0 < a < 1 (D) a = 3 1 or a = – 3

   ›Reveal solutionSolution

   The direction cosines of a line must satisfy $l^2 + m^2 + n^2 = 1$. Given all three are equal to $a$, we get $3a^2 = 1$, so $a = \pm \frac{1}{\sqrt{3}}$. The correct option is (D).

   Direction cosines are the cosines of the angles a line makes with the coordinate axes. A fundamental property — and the key to this problem — is that the sum of their squares is always exactly 1. This isn't arbitrary; it follows from the fact that a unit vector along the line has components equal to the direction cosines, and its magnitude must be 1.

   Here, all three direction cosines are given as the same number $a$. That immediately tells us the line makes equal angles with all three axes — it's symmetrically oriented, like the body diagonal of a cube. But the value of $a$ isn't free; it's forced by the square-sum rule.

   Let's work it out.

   1. Write the condition. If $l, m, n$ are direction cosines, then

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

   This is non-negotiable — it's the defining constraint.

   2. Substitute the given values. Here $l = m = n = a$. So

   $a^2 + a^2 + a^2 = 1 \quad \Rightarrow \quad 3a^2 = 1.$

   3. Solve for $a$.

   $a^2 = \frac{1}{3} \quad \Rightarrow \quad a = \pm \frac{1}{\sqrt{3}}.$

   4. Check the options.
      • (A) $a > 0$ — false, because $a$ can be negative.
      • (B) $a = 1$ or $a = -1$ — false; those would give $1+1+1=3 \neq 1$.
      • (C) $0 < a < 1$ — false, since $a$ can be negative.
      • (D) $a = \frac{1}{\sqrt{3}}$ or $a = -\frac{1}{\sqrt{3}}$ — correct.

   Watch out

   A common mistake is to forget that direction cosines can be negative. A line can make an obtuse angle with an axis, giving a negative cosine. So $a$ is not necessarily positive.

   Tip

   If you ever see "direction cosines are equal", immediately think of the cube's diagonal. The cosines are $\pm \frac{1}{\sqrt{3}}$, and the sign depends on which octant the line points into.

   ✓Final answer

   The correct option is (D): $a = \frac{1}{\sqrt{3}}$ or $a = -\frac{1}{\sqrt{3}}$.

3. CBSE 20231 markMCQ
   Q.Direction cosines of the line $\frac{x-1}{2} = \frac{1-y}{3} = \frac{2z-1}{12}$ are : (A) $\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$ (B) $\frac{2}{\sqrt{157}}, \frac{-3}{\sqrt{157}}, \frac{12}{\sqrt{157}}$ (C) $\frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}$ (D) $\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}$

   ›Reveal solutionSolution

   The line has direction ratios $(2, -3, 6)$, giving direction cosines $\left(\tfrac{2}{7}, -\tfrac{3}{7}, \tfrac{6}{7}\right)$. This exact set is not among the printed options — see the note below.

   Direction cosines are the direction ratios scaled so their squares sum to $1$. First put the line in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.

   Rewrite the $y$-term: $\dfrac{1-y}{3} = \dfrac{-(y-1)}{3} = \dfrac{y-1}{-3}$, so $b = -3$.

   Rewrite the $z$-term: $\dfrac{2z-1}{12} = \dfrac{2\left(z - \tfrac{1}{2}\right)}{12} = \dfrac{z - \tfrac{1}{2}}{6}$, so $c = 6$.

   Standard form:

   $\frac{x-1}{2} = \frac{y-1}{-3} = \frac{z - \tfrac{1}{2}}{6}.$

   Direction ratios: $(2, -3, 6)$.

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

   Direction cosines:

   $\left(\frac{2}{7},\ -\frac{3}{7},\ \frac{6}{7}\right), \qquad \left(\tfrac{2}{7}\right)^2 + \left(-\tfrac{3}{7}\right)^2 + \left(\tfrac{6}{7}\right)^2 = \frac{4 + 9 + 36}{49} = 1.$

   A parametrisation confirms the sign of the $z$-component: with common value $t$, $z = \tfrac{1}{2} + 6t$, so the $z$-direction component is $+6$.

   Note

   The only two valid direction-cosine sets for this line are $\left(\tfrac{2}{7}, -\tfrac{3}{7}, \tfrac{6}{7}\right)$ and its negative $\left(-\tfrac{2}{7}, \tfrac{3}{7}, -\tfrac{6}{7}\right)$. None of the four printed options states either set exactly: option (C), $\left(\tfrac{2}{7}, -\tfrac{3}{7}, -\tfrac{6}{7}\right)$, matches on $x$ and $y$ but has the wrong sign on the $z$-cosine. The printed options contain a sign error.

   ✓Final answer

   The direction cosines of the line are $\left(\dfrac{2}{7},\ -\dfrac{3}{7},\ \dfrac{6}{7}\right)$. As printed, no option is exactly correct — the nearest, option (C), has an incorrect sign on the $z$-component.

4. CBSE 20191 mark
   Q.Find the direction cosines of a line which makes equal angles with the coordinate axes.

   (OR)

   Find the cartesian equation of the line which passes through the point with position vector $2\hat{i} - \hat{j} + 4\hat{k}$ and is in the direction of the vector $\hat{i} + \hat{j} - 2\hat{k}$. Find the direction cosines of a line which makes equal angles with the coordinate axes.

   (OR)

   A line passes through the point with position vector $2\hat{i} - \hat{j} + 4\hat{k}$ and is in the direction of the vector $\hat{i} + \hat{j} - 2\hat{k}$. Find the equation of the line in cartesian form.

   ›Reveal solutionSolution

   Part (a): equal angles give direction cosines $\left(\pm\tfrac{1}{\sqrt3},\pm\tfrac{1}{\sqrt3},\pm\tfrac{1}{\sqrt3}\right)$. Parts (b) & (c): the line through $(2,-1,4)$ along $(1,1,-2)$ has cartesian form $\dfrac{x-2}{1}=\dfrac{y+1}{1}=\dfrac{z-4}{-2}$.

   Part (a)

   Direction cosines $l,m,n$ are the cosines of the angles the line makes with the $x,y,z$ axes and always satisfy $l^2+m^2+n^2=1$.

   1. Equal angles. Equal angles with all three axes means $l=m=n=k$.
   2. Apply the identity. $k^2+k^2+k^2=1\Rightarrow 3k^2=1\Rightarrow k^2=\tfrac13$.
   3. Solve. $k=\pm\dfrac{1}{\sqrt3}$.

   The signs must all be $+$ or all $-$ (choosing a direction along the line).

   Watch out

   The direction cosines are not $(1,1,1)$; each equal angle is $\cos^{-1}\!\tfrac{1}{\sqrt3}\approx 54.7^\circ$, not $45^\circ$.

   ✓Final answer

   Direction cosines: $\left(\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3}\right)$ or $\left(-\tfrac{1}{\sqrt3},-\tfrac{1}{\sqrt3},-\tfrac{1}{\sqrt3}\right)$.

   Part (b)

   This part asks for the cartesian equation of the line through the point with position vector $2\hat i-\hat j+4\hat k$ in the direction $\hat i+\hat j-2\hat k$ (and restates the equal-angle question of Part (a)).

   1. Read off the data. Point $(x_1,y_1,z_1)=(2,-1,4)$; direction ratios $(a,b,c)=(1,1,-2)$.
   2. Cartesian form. $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$, so

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

   For the restated equal-angle sub-part, the direction cosines are $\pm\tfrac{1}{\sqrt3}$ each (see Part (a)).

   ✓Final answer

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

   Part (c)

   Same data: point $(2,-1,4)$ and direction ratios $(1,1,-2)$.

   1. Substitute into the symmetric form.

   $\frac{x-2}{1}=\frac{y-(-1)}{1}=\frac{z-4}{-2}=\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-4}{-2}.$

   Tip

   The equivalent vector form is $\vec r=(2\hat i-\hat j+4\hat k)+\lambda(\hat i+\hat j-2\hat k)$.

   ✓Final answer

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

5. CBSE 2026Set 65/2/11 markMCQ
   Q.Direction cosines of the line given by equations $\frac{2x-1}{4} = \frac{1-y}{3} = \frac{-z}{6}$ are (A) $2, -3, -6$ (B) $\frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}$ (C) $\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}$ (D) $\frac{4}{\sqrt{61}}, \frac{-3}{\sqrt{61}}, \frac{-6}{\sqrt{61}}$

   ›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

   1. 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:

   $\frac{x - 1/2}{2} = \frac{y - 1}{-3} = \frac{z - 0}{-6}$

   > [!WARNING]
   > A common mistake is to directly take $(4, 3, 6)$ or $(4, -3, -6)$ as direction ratios. This is incorrect because the numerators were not in the standard $(x-x_1)$, $(y-y_1)$, $(z-z_1)$ form. Always ensure the coefficient of $x, y, z$ in the numerator is $+1$.
   

   2. Identify the direction ratios.

   From the standard form $\frac{x - 1/2}{2} = \frac{y - 1}{-3} = \frac{z - 0}{-6}$, the direction ratios $(a, b, c)$ are the denominators.

   So, $a = 2$, $b = -3$, $c = -6$.

   3. Calculate the magnitude of the direction vector.

      The magnitude is $\sqrt{a^2 + b^2 + c^2}$.

      Magnitude $= \sqrt{(2)^2 + (-3)^2 + (-6)^2}$

      Magnitude $= \sqrt{4 + 9 + 36}$

      Magnitude $= \sqrt{49}$

      Magnitude $= 7$.

   4. Calculate the direction cosines.

      The direction cosines $(l, m, n)$ are obtained by dividing each direction ratio by the magnitude.

      $l = \frac{a}{\text{Magnitude}} = \frac{2}{7}$

      $m = \frac{b}{\text{Magnitude}} = \frac{-3}{7}$

      $n = \frac{c}{\text{Magnitude}} = \frac{-6}{7}$

      So, the direction cosines are $\left(\frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}\right)$.

      Comparing this with the given options, option (B) matches our result.

   ✓Final answer

   The direction cosines of the given line are $\boxed{\frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}}$.

6. CBSE 2026Set 65/2/11 markMCQ
   Q.Assertion (A): A line can have direction cosines $< 1, 1, 1 >$. Reason (R): $\cos \theta = 1$ is possible for $\theta = 0$. (A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

   ›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.

   1. 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.

   2. 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.

   3. Does Reason (R) explain Assertion (A)?

      The Reason talks about a single cosine being 1, but the Assertion claims a triple of three 1’s can be direction cosines. Even if $\cos\theta = 1$ is possible, that doesn’t make $<1,1,1>$ valid — because the sum-of-squares condition fails. So (R) is not the correct explanation of (A).

   Watch out

   A common mistake is to think that because each individual number 1 is a possible cosine (for angle $0^\circ$), the triple must be valid. But direction cosines are not independent — they must satisfy $l^2+m^2+n^2=1$. Three 1’s break that rule.

   Thus, Assertion (A) is false, Reason (R) is true, and (R) does not explain (A). That matches option (D).

   ✓Final answer

   The correct option is (D).

7. CBSE 2026Set CX1 mark
   Q.If a line makes $90^{\circ}$, $60^{\circ}$ and $30^{\circ}$ with $x$, $y$ and $z$-axes in the positive direction respectively, then find direction cosines.

   ›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 $l=\cos\alpha,\ m=\cos\beta,\ n=\cos\gamma$.

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

   Check: $l^{2}+m^{2}+n^{2}=0+\dfrac14+\dfrac34=1$ ✓ (as required for direction cosines).

   ✓Final answer

   Direction cosines $=\left(0,\ \dfrac{1}{2},\ \dfrac{\sqrt3}{2}\right)$.

8. CBSE 2026Set A1 markMCQ
   Q.The direction ratios of a straight line are $2, 6, -3$. Then its direction cosines are

   (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

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

   So the direction cosines are $\dfrac{2}{7}, \dfrac{6}{7}, \dfrac{-3}{7}$.

   ✓Final answer

   (c) $\frac{2}{7}, \frac{6}{7}, \frac{-3}{7}$.

9. CBSE 2026Set A1 markMCQ
   Q.If a line makes angles $\alpha$, $\beta$ and $\gamma$ with the positive directions of $x$, $y$ and $z$ axes respectively, then

   (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.

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

   (Equivalently $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$, not $1$, so option (d) is wrong.)

   ✓Final answer

   (a) $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

10. CBSE 2026Set ANNUAL1 markMCQ
    Q.If a line makes angles of $30^{\circ}$ and $45^{\circ}$ with $X$-axis and $Y$-axis respectively, then what is the angle made by it with $Z$-axis?

    (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

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

    This is negative, which is impossible for any real $\cos^2\gamma\ge0$. Hence no real angle $\gamma$ satisfies the given combination of $30^{\circ}$ and $45^{\circ}$ with the other two axes — such a line does not exist, so the angle with the $Z$-axis cannot be determined.

    ✓Final answer

    The correct option is (d) Cannot be determined (the given pair of angles is inconsistent with $l^2+m^2+n^2=1$).

11. CBSE 2026Set ANNUAL1 mark
    Q.Find the direction cosines of the line passing through the two points $(-2, 4, -5)$ and $(1, 2, 3)$.

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

    Direction cosines $=\left(\dfrac{3}{\sqrt{77}},\dfrac{-2}{\sqrt{77}},\dfrac{8}{\sqrt{77}}\right)$.

    ✓Final answer

    Direction cosines $=\left(\dfrac{3}{\sqrt{77}},\dfrac{-2}{\sqrt{77}},\dfrac{8}{\sqrt{77}}\right)$.

12. CBSE 2026Set ANNUAL1 markMCQ
    Q.If a line makes angles $\alpha, \beta, \gamma$ with coordinate axes then $\sin^2\alpha + \sin^2\beta + \sin^2\gamma =$

    (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:

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

    So $\sin^2\alpha+\sin^2\beta+\sin^2\gamma = (1-\cos^2\alpha)+(1-\cos^2\beta)+(1-\cos^2\gamma) = 3 - 1 = 2$.

    ✓Final answer

    (a) $2$.