Q.If $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}+5\hat{j}$, $3\hat{i}+2\hat{j}-3\hat{k}$ and $\hat{i}-6\hat{j}-\hat{k}$ are the position vectors of points A, B, C and D respectively, then find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$. Deduce that $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.

Showing the 12 most recent of 14 on this concept.

1. GUJCET 2024Set 131 markMCQ
   Q.The angle '$\theta$' between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ is __________. (A) $\sin^{-1}\left(-\frac{1}{3}\right)$ (B) $\cos^{-1}\left(-\frac{1}{3}\right)$ (C) $\sin^{-1}\frac{1}{3}$ (D) $\cos^{-1}\frac{1}{3}$

   ›Reveal solutionSolution

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

   Steps. $\vec a=(1,-1,1)$, $\vec b=(1,1,-1)$.

   $\vec a\cdot\vec b = 1-1-1 = -1,\quad |\vec a|=|\vec b|=\sqrt{3}.$

   $\cos\theta = \frac{-1}{\sqrt3\cdot\sqrt3} = -\frac{1}{3} \ \Rightarrow\ \theta=\cos^{-1}\!\left(-\tfrac{1}{3}\right).$

   ✓Final answer

   Option (B) $\cos^{-1}\left(-\frac{1}{3}\right)$

   ANSWER: (B)

2. GUJCET 2023Set 091 markMCQ
   Q.Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$ and $\vec{a}\times\vec{b}$ is a unit vector, then the angle between $\vec{a}$ and $\vec{b}$ is : (A) $\frac{\pi}{2}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{6}$ (D) $\frac{\pi}{4}$

   ›Reveal solutionSolution

   The magnitude of a cross product is $|\vec a||\vec b|\sin\theta$; setting it to 1 fixes the angle.

   Concept: $|\vec a\times\vec b| = |\vec a|\,|\vec b|\sin\theta$. Here the cross product is a unit vector, so its magnitude is $1$.

   $3\cdot\frac{\sqrt2}{3}\cdot\sin\theta = 1 \;\Rightarrow\; \sqrt2\,\sin\theta = 1 \;\Rightarrow\; \sin\theta = \frac{1}{\sqrt2}.$

   Hence $\theta = \dfrac{\pi}{4}$.

   ✓Final answer

   (D) $\dfrac{\pi}{4}$

   ANSWER: (D)

3. GUJCET 2022Set 081 markMCQ
   Q.Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$. If $\vec{a}\times\vec{b}$ is a unit vector, then the angle between $\vec{a}$ and $\vec{b}$ is ______. (A) $\frac{\pi}{2}$ (B) $\frac{\pi}{4}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{6}$

   ›Reveal solutionSolution

   $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta=1$ solves to $\theta=\tfrac\pi4$.

   Concept. $|\vec a\times\vec b|=3\cdot\dfrac{\sqrt2}{3}\sin\theta=\sqrt2\sin\theta$. Setting this $=1$: $\sin\theta=\dfrac{1}{\sqrt2}$, so $\theta=\dfrac{\pi}{4}$.

   ✓Final answer

   (B) $\frac{\pi}{4}$

   ANSWER: (B)

4. GUJCET 2021Set 151 markMCQ
   Q.Let the vector $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is $\underline{\hspace{2cm}}$ (A) $\frac{\pi}{6}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   Set the magnitude of the cross product to 1 and solve for the angle.

   Concept. $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta$.

   Solution. $3\cdot\dfrac{\sqrt2}{3}\sin\theta=\sqrt2\sin\theta=1\Rightarrow\sin\theta=\dfrac{1}{\sqrt2}\Rightarrow\theta=\dfrac{\pi}{4}$.

   ✓Final answer

   (C) $\frac{\pi}{4}$

   ANSWER: (C)

5. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.The angle between the vectors $\hat i-\hat j-\hat k$ and $\hat i-\hat j+\hat k$ is ____.

   (a) $\sin^{-1}\dfrac{2\sqrt2}{3}$

   (b) $\cos^{-1}\left(-\dfrac13\right)$

   (c) $\cos^{-1}\dfrac{2\sqrt2}{3}$

   (d) $\sin^{-1}\left(-\dfrac13\right)$

   ›Reveal solutionSolution

   Find $\cos\theta$ from the dot product, then express the same acute angle using $\sin\theta$.

   Let $\vec a=\hat i-\hat j-\hat k$, $\vec b=\hat i-\hat j+\hat k$. $\vec a\cdot\vec b=1+1-1=1$. $|\vec a|=|\vec b|=\sqrt3$.

   $\cos\theta=\dfrac{1}{3}\Rightarrow\sin\theta=\sqrt{1-\frac19}=\dfrac{2\sqrt2}{3}$ (positive since $\theta$ is acute). So $\theta=\sin^{-1}\dfrac{2\sqrt2}{3}$ (equivalently $\cos^{-1}\frac13$).

   ✓Final answer

   The correct option is (a) $\sin^{-1}\dfrac{2\sqrt2}{3}$.

6. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.If $\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} - 5\hat{k}$, then the measure of the angle between the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ = ____.

   (a) $0$

   (b) $\pi$

   (c) $\dfrac{\pi}{2}$

   (d) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   $(\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2$; if this is zero the vectors are perpendicular.

   $|\vec a|^2=25+1+9=35$, $|\vec b|^2=1+9+25=35$.

   $(\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2=35-35=0$, so the vectors are perpendicular: angle $=\dfrac{\pi}{2}$.

   ✓Final answer

   The correct option is (c) $\dfrac{\pi}{2}$.

7. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$, then for $\theta$ = ____, $|\vec{a}\cdot\vec{b}| = |\vec{a}\times\vec{b}|$.

   (a) $0$

   (b) $\dfrac{\pi}{2}$

   (c) $\dfrac{\pi}{4}$

   (d) $\pi$

   ›Reveal solutionSolution

   $|\vec a\cdot\vec b|=|\vec a||\vec b||\cos\theta|$ and $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta$; equating gives $\tan\theta=1$.

   $|\cos\theta|=\sin\theta \Rightarrow \tan\theta=1$ (for $\theta\in[0,\pi]$ with $\sin\theta\ge0$), so $\theta=\dfrac{\pi}{4}$.

   ✓Final answer

   The correct option is (c) $\dfrac{\pi}{4}$.

8. GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
   Q.The angle between the vectors $\vec{a} = 6\hat{i} + 2\hat{j} - 8\hat{k}$ and $\vec{b} = 4\hat{i} - 4\hat{j} + 2\hat{k}$ is ______.

   (a) $\frac{\pi}{3}$

   (b) $\frac{\pi}{2}$

   (c) $\frac{\pi}{4}$

   (d) $0$

   ›Reveal solutionSolution

   A zero dot product means the vectors are perpendicular.

   $\vec a\cdot\vec b=6(4)+2(-4)+(-8)(2)=24-8-16=0$.

   Since $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0$ and neither vector is zero, $\theta=\frac{\pi}{2}$.

   ✓Final answer

   The correct option is (b) $\frac{\pi}{2}$.

9. GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
   Q.If $|\vec{a}| = 10$, $|\vec{b}| = 2$ and $\vec{a} \cdot \vec{b} = 12$, then value of $|\vec{a} \times \vec{b}|$ is ___.

   (a) $5$

   (b) $10$

   (c) $16$

   (d) $14$

   ›Reveal solutionSolution

   Use the identity $|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2$.

   $|\vec a\times\vec b|^2 = (10)^2(2)^2 - (12)^2 = 400 - 144 = 256.$

   $|\vec a\times\vec b| = 16.$

   ✓Final answer

   (c) $16$.

10. GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
    Q.The angle '$\theta$' between the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is ___.

    (a) $\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$

    (b) $\cos^{-1}\left(-\frac{1}{3}\right)$

    (c) $-\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$

    (d) None of the above

    ›Reveal solutionSolution

    Find $\cos\theta$ from the dot product, then express as an $\sin^{-1}$ angle.

    $\vec a\cdot\vec b = 1-1+1 = 1$, $|\vec a| = |\vec b| = \sqrt3.$

    $\cos\theta = \dfrac{1}{\sqrt3\cdot\sqrt3} = \dfrac13.$

    Then $\sin\theta = \sqrt{1-\tfrac19} = \dfrac{2\sqrt2}{3}$ (acute angle), so $\theta = \sin^{-1}\!\left(\dfrac{2\sqrt2}{3}\right).$

    ✓Final answer

    (a) $\sin^{-1}\left(\tfrac{2\sqrt{2}}{3}\right)$.

11. GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
    Q.Measure of the angle between the vectors $\vec a=\hat\imath-\hat\jmath+\hat k$ and $\vec b=\hat\imath+\hat\jmath+\hat k$ is ___.

    (a) $\cos^{-1}\dfrac{1}{\sqrt3}$

    (b) $\pi-\cos^{-1}\dfrac13$

    (c) $\sin^{-1}\dfrac{2\sqrt2}{3}$

    (d) $\sin^{-1}\dfrac13$

    ›Reveal solutionSolution

    Use $\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}$, then convert to the equivalent $\sin^{-1}$ form to match the given options.

    $\vec a=\hat\imath-\hat\jmath+\hat k,\ \vec b=\hat\imath+\hat\jmath+\hat k$. $\vec a\cdot\vec b=1-1+1=1$. $|\vec a|=|\vec b|=\sqrt3$.

    $\cos\theta=\dfrac1{\sqrt3\cdot\sqrt3}=\dfrac13$, so $\theta=\cos^{-1}\dfrac13$ (an acute angle, since $\cos\theta>0$).

    Since $\theta$ is acute, $\sin\theta=\sqrt{1-\tfrac19}=\sqrt{\tfrac89}=\dfrac{2\sqrt2}3$, so $\theta=\sin^{-1}\dfrac{2\sqrt2}{3}$ equally.

    ✓Final answer

    (c) $\sin^{-1}\dfrac{2\sqrt2}{3}$.

12. GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
    Q.Let $|\vec x|=|\vec y|=|\vec x+\vec y|=1$ and if measure of the angle between $\vec x$ and $\vec y$ is $\alpha$, then $\sin\alpha=$ ___

    (a) $-\dfrac{\sqrt3}{2}$

    (b) $\dfrac{\sqrt3}{2}$

    (c) $-\dfrac12$

    (d) 1

    ›Reveal solutionSolution

    Square the condition $|\vec x+\vec y|=1$ to extract $\vec x\cdot\vec y$, hence $\cos\alpha$, then find $\sin\alpha$.

    $|\vec x+\vec y|^2=|\vec x|^2+|\vec y|^2+2\vec x\cdot\vec y=1+1+2\vec x\cdot\vec y=1 \Rightarrow \vec x\cdot\vec y=-\dfrac12$.

    $\cos\alpha=\dfrac{\vec x\cdot\vec y}{|\vec x||\vec y|}=-\dfrac12$, so $\alpha=\dfrac{2\pi}{3}$ (120°).

    $\sin\alpha=\sin\dfrac{2\pi}3=\dfrac{\sqrt3}{2}$.

    ✓Final answer

    (b) $\dfrac{\sqrt3}{2}$.