Q.Find a vector $\vec{r}$ equally inclined to the three axes and whose magnitude is $3\sqrt{3}$ units.

Showing the 12 most recent of 152 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.If $|\vec{a}| = 8$, $|\vec{b}| = 3$ and $|\vec{a} \times \vec{b}| = 12$, then the value of $|\vec{a} \cdot \vec{b}|$ is (A) $6\sqrt{3}$ (B) $8\sqrt{3}$ (C) $12\sqrt{3}$ (D) None of these

   ›Reveal solutionSolution

   The key is the identity $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$. Substituting the given magnitudes gives $144 + (\vec{a} \cdot \vec{b})^2 = 576$, so $|\vec{a} \cdot \vec{b}| = \sqrt{432} = 12\sqrt{3}$. The correct option is (C).

   The problem gives you the magnitudes of two vectors and the magnitude of their cross product, and asks for the magnitude of their dot product. This is a classic setup — it tests a single, powerful relationship that ties the dot product and cross product together.

   The core idea: The dot product depends on $\cos\theta$, and the cross product depends on $\sin\theta$, where $\theta$ is the angle between the vectors. Since $|\vec{a}|$ and $|\vec{b}|$ are known, you can use the identity $\sin^2\theta + \cos^2\theta = 1$ to eliminate $\theta$ and directly connect the two products.

   1. Write the definitions:

      • $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$
      • $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$

      Here $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, with $0 \le \theta \le \pi$.

   2. Square both equations:

      • $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2\theta$
      • $(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2\theta$

   3. Add them together:

   $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 (\sin^2\theta + \cos^2\theta) = |\vec{a}|^2 |\vec{b}|^2$

   This is the identity you need. It holds for any two vectors in 3D space.

   $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$

   4. Plug in the given numbers:

      • $|\vec{a}| = 8$, $|\vec{b}| = 3$, so $|\vec{a}|^2 |\vec{b}|^2 = 64 \times 9 = 576$
      • $|\vec{a} \times \vec{b}| = 12$, so $|\vec{a} \times \vec{b}|^2 = 144$

      Therefore: …

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2. 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:

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

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3. CBSE 2026Set 65/2/11 markMCQ
   Q.For two vectors $\vec{a}$ and $\vec{b}$: Assertion (A): $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$ Reason (R): $|\vec{a} \times \vec{b}| = (\vec{a} \cdot \vec{b}) \tan \theta$, $\left(\theta \neq \frac{\pi}{2}\right)$ (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

   Assertion (A) is a fundamental identity relating the magnitudes of the cross product and dot product, which is true. Reason (R) is also a true relationship between the magnitudes of the cross product and dot product, but it does not explain Assertion (A). The correct option is (B).

   To evaluate this assertion-reason question, we need to understand the definitions of the dot product and cross product of two vectors and the geometric meaning of the angle between them. Both the dot product and the cross product are fundamental operations in vector algebra, and their properties are frequently tested.

   The dot product (or scalar product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:

   $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

   where $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of vectors $\vec{a}$ and $\vec{b}$ respectively, and $\theta$ is the angle between them ($0 \le \theta \le \pi$). The result is a scalar.

   The cross product (or vector product) of two vectors $\vec{a}$ and $\vec{b}$ results in a vector perpendicular to both $\vec{a}$ and $\vec{b}$. Its magnitude is defined as:

   $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$

   where $\theta$ is again the angle between $\vec{a}$ and $\vec{b}$. The direction of $\vec{a} \times \vec{b}$ is given by the right-hand rule.

   Now, let's evaluate the Assertion and Reason.

   1. Evaluate Assertion (A):

      The assertion states: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$.

      Let's substitute the definitions of $|\vec{a} \times \vec{b}|$ and $(\vec{a} \cdot \vec{b})$ into the left-hand side (LHS) of the equation.

      LHS $= (|\vec{a}| |\vec{b}| \sin \theta)^2 + (|\vec{a}| |\vec{b}| \cos \theta)^2$

      LHS $= |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$

      We can factor out $|\vec{a}|^2 |\vec{b}|^2$:

      LHS $= |\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta)$

      Important

      Recall the fundamental trigonometric identity: $\sin^2 \theta + \cos^2 \theta = 1$.

      Using this identity:

      LHS $= |\vec{a}|^2 |\vec{b}|^2 (1)$

      LHS $= |\vec{a}|^2 |\vec{b}|^2$

      This matches the right-hand side (RHS) of the assertion.

      Therefore, Assertion (A) is True. This identity is often known as Lagrange's Identity for vectors.

   2. Evaluate Reason (R):

      The reason states: $|\vec{a} \times \vec{b}| = (\vec{a} \cdot \vec{b}) \tan \theta$, $\left(\theta \neq \frac{\pi}{2}\right)$.

      Let's substitute the definitions of $|\vec{a} \times \vec{b}|$ and $(\vec{a} \cdot \vec{b})$ into this equation.

      LHS: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$

      RHS: $(\vec{a} \cdot \vec{b}) \tan \theta = (|\vec{a}| |\vec{b}| \cos \theta) \tan \theta$

      Recall the definition of $\tan \theta$: $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

      Substitute this into the RHS:

      RHS $= (|\vec{a}| |\vec{b}| \cos \theta) \left(\frac{\sin \theta}{\cos \theta}\right)$

      Since $\theta \neq \frac{\pi}{2}$, $\cos \theta \neq 0$, so we can cancel $\cos \theta$: …

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4. 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)? …

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

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

   …

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6. 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

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

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7. 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.

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

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8. 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

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

   …

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9. CBSE 2026Set ANNUAL1 mark
   Q.Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$ respectively and $\vec{a}.\vec{b} = \sqrt{6}$.

   ›Reveal solutionSolution

   Use $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ to solve for $\theta$.

   $\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \dfrac{\sqrt6}{\sqrt3\cdot2} = \dfrac{\sqrt6}{2\sqrt3}=\dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}$

   …

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

    …

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11. 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:

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

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.The angle between the vectors $\hat{i} + 3\hat{j} + 3\hat{k}$ and $3\hat{i} - 2\hat{j} + \hat{k}$ is:

    (a) $0°$

    (b) $45°$

    (c) $60°$

    (d) $90°$

    ›Reveal solutionSolution

    The two vectors have zero dot product, so they are perpendicular.

    Let $\vec{a}=\hat{i}+3\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+\hat{k}$.

    $\vec{a}\cdot\vec{b}=1(3)+3(-2)+3(1)=3-6+3=0$

    …

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