Q.(a) Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors $\vec{a} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k}$. Determine the angle formed between the kite strings. Assume there is no slack in the strings.

Showing the 12 most recent of 106 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.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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3. CBSE 2026Set V11 markMCQ
   Q.If $\vec{a}$ is a nonzero vector of magnitude $a$ and $\lambda$, a nonzero scalar then $\lambda\vec{a}$ is a unit vector if

   (a) $\lambda=1$

   (b) $\lambda=-1$

   (c) $a=|\lambda|$

   (d) $a=\frac{1}{|\lambda|}$

   ›Reveal solutionSolution

   Requiring $|\lambda\vec a|=1$ gives $|\lambda|a=1$, i.e. $a=\dfrac{1}{|\lambda|}$; answer (d).

   The magnitude of $\lambda\vec a$ is

   $|\lambda\vec a|=|\lambda|\,|\vec a|=|\lambda|\,a.$ …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The unit vector in the direction of the vector $\vec{a} = \hat{i}+\hat{j}+2\hat{k}$ is

   (a) $\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{5}}$

   (b) $\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}$

   (c) $\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}$

   (d) $\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}$

   ›Reveal solutionSolution

   A unit vector along $\vec a$ is $\vec a/|\vec a|$.

   $|\vec a| = \sqrt{1^2+1^2+2^2}=\sqrt6$.

   Unit vector $=\dfrac{\hat i+\hat j+2\hat k}{\sqrt6}$.

   …

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5. 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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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\vec{a} = 2\hat{i} - 7\hat{j} - 3\hat{k}$ then $\hat{a} =$

   (a) $\dfrac{2\hat{i}-7\hat{j}-3\hat{k}}{\sqrt{62}}$

   (b) $2\hat{i}-7\hat{j}-3\hat{k}$

   (c) $\dfrac{1}{\sqrt{62}}$

   (d) None of these

   ›Reveal solutionSolution

   A unit vector in the direction of $\vec a$ is $\hat a = \dfrac{\vec a}{|\vec a|}$.

   $|\vec a| = \sqrt{2^2+(-7)^2+(-3)^2} = \sqrt{4+49+9} = \sqrt{62}$.

   …

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7. 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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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.If vector a . vector b = √3 |vector a × vector b| then angle between vector a and vector b is:

   (a) π/2

   (b) π/6

   (c) π/4

   (d) π/3

   ›Reveal solutionSolution

   Writing the dot and cross product magnitudes in terms of $\cos\theta$ and $\sin\theta$ turns the given condition into $\tan\theta = \frac{1}{\sqrt3}$.

   We know:

   $\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta, \qquad |\vec a \times \vec b| = |\vec a||\vec b|\sin\theta$

   Given $\vec a \cdot \vec b = \sqrt3\, |\vec a \times \vec b|$:

   $|\vec a||\vec b|\cos\theta = \sqrt3\,|\vec a||\vec b|\sin\theta$

   …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.If |\vec a| = 1, |\vec b| = 2 and \vec a \cdot \vec b = 1, then angle between \vec a and \vec b is:

   (a) \pi/2

   (b) \pi/6

   (c) \pi/3

   (d) \pi/4

   ›Reveal solutionSolution

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

   Working:

   $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ …

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10. CBSE 2026Set ANNUAL1 markMCQ
    Q.Dot Product of two vectors is defined as

    (a) a·b = |a||b| sin θ

    (b) a·b = |a||b| cos θ

    (c) a·b = |a||b| sin θ n̂

    (d) None of the above

    ›Reveal solutionSolution

    The dot (scalar) product uses cosine of the angle between the vectors; the cross (vector) product uses sine.

    For two vectors $\vec a$ and $\vec b$ with angle $\theta$ between them, the dot product is defined as

    $\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta,$ …

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11. CBSE 2026Set ANNUAL1 mark
    Q.Find the unit vector in the direction of the vector $\vec{a} = \hat{i}+\hat{j}+2\hat{k}$.

    ›Reveal solutionSolution

    Compute the magnitude of $\vec a$, then divide the vector by its magnitude to get the unit vector.

    Given $\vec a = \hat{i}+\hat{j}+2\hat{k}$.

    Step 1: Find the magnitude.

    $|\vec a| = \sqrt{1^2+1^2+2^2} = \sqrt{1+1+4} = \sqrt{6}$

    Step 2: Divide by the magnitude.

    $\hat{a} = \frac{\vec a}{|\vec a|} = \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt6}$

    $\hat{a} = \frac{1}{\sqrt6}\hat{i}+\frac{1}{\sqrt6}\hat{j}+\frac{2}{\sqrt6}\hat{k}$

    …

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.If the scalar product of two vectors is zero, then the angle between them is:

    (a) 45°

    (b) 0°

    (c) 180°

    (d) 90°

    ›Reveal solutionSolution

    A.B = AB*cos(theta) is zero only when cos(theta) = 0, i.e., when the vectors are perpendicular, theta = 90 degrees.

    The scalar product (dot product) of two vectors A and B is defined as

    A.B = |A||B|*cos(theta)

    where theta is the angle between the two vectors.

    For the scalar product to be zero (assuming neither vector is itself the zero vector), we need

    cos(theta) = 0

    …

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