Q.If $\vec{a}$ is a nonzero vector with magnitude $a$, and $\lambda$ is a nonzero scalar, then for what value of $\lambda$ does $\lambda\vec{a}$ become a unit vector?

Showing the 12 most recent of 40 on this concept.

1. 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

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

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $=\dfrac{\hat i+\hat j+2\hat k}{\sqrt6}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $\hat a = \dfrac{\vec a}{|\vec a|}$.

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

   …

   🔒 Start your 14-day free trial to see the full solution →
4. 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.

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

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

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $\vec a=\hat i+7\hat j+4\hat k$ and $\vec b=3\hat i+\hat j+\hat k$, then what is the unit vector in the direction of $\vec a-\vec b$?

   (i) $\dfrac{1}{7}(2\hat i+6\hat j-3\hat k)$

   (ii) $\dfrac{1}{7}(-2\hat i+6\hat j+3\hat k)$

   (iii) $\dfrac{1}{7}(-2\hat i-6\hat j+3\hat k)$

   (iv) $\dfrac{1}{7}(2\hat i-6\hat j+3\hat k)$

   ›Reveal solutionSolution

   Find $\vec a - \vec b$, its magnitude, then divide by the magnitude.

   $\vec a - \vec b = (1-3)\hat i + (7-1)\hat j + (4-1)\hat k = -2\hat i + 6\hat j + 3\hat k$

   Magnitude:

   $|\vec a-\vec b| = \sqrt{(-2)^2+6^2+3^2} = \sqrt{4+36+9} = \sqrt{49} = 7$

   Unit vector: …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $\vec{a}$ is a non-zero vector of magnitude '$a$' and '$\lambda$' is a non-zero scalar, then $\lambda\vec{a}$ is a unit vector if OR The value of $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$ is

   (a) 0

   (b) -1

   (c) 1

   (d) 3

   (a) $\lambda=1$

   (b) $\lambda=-1$

   (c) $a=|\lambda|$

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

   ›Reveal solutionSolution

   A vector is a unit vector when its magnitude equals $1$; use $|\lambda\vec a|=|\lambda||\vec a|$.

   Given $|\vec a|=a\ (\ne0)$ and scalar $\lambda\ (\ne0)$, the magnitude of $\lambda\vec a$ is

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

   For $\lambda\vec a$ to be a unit vector we need $|\lambda\vec a|=1$:

   $|\lambda|\,a=1\ \Rightarrow\ a=\frac{1}{|\lambda|}.$

   Checking the options: (a) $\lambda=1$ and (b) $\lambda=-1$ force a specific $\lambda$ but ignore $a$; (c) $a=|\lambda|$ is the reciprocal of the correct relation. Only (d) $a=\frac{1}{|\lambda|}$ works.

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The unit vector in the direction of the vector $\vec a=\hat i+\hat j+2\hat k$ is

   (a) $\dfrac{1}{2}\hat i+\dfrac{1}{2}\hat j+\hat k$

   (b) $\dfrac{1}{\sqrt 3}\hat i+\dfrac{1}{\sqrt 3}\hat j+\dfrac{2}{\sqrt 3}\hat k$

   (c) $\dfrac{1}{\sqrt 5}\hat i+\dfrac{1}{\sqrt 5}\hat j+\dfrac{2}{\sqrt 5}\hat k$

   (d) $\dfrac{1}{\sqrt 6}\hat i+\dfrac{1}{\sqrt 6}\hat j+\dfrac{2}{\sqrt 6}\hat k$

   ›Reveal solutionSolution

   Unit vector $=\vec a/|\vec a|$; here $|\vec a|=\sqrt6$.

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

   $|\vec a|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt6$

   The unit vector in the direction of $\vec a$ is: …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set ANNUAL1 markMCQ
   Q.If a⃗ is a non-zero vector of magnitude 'a' and λ is a non-zero scalar, then λa⃗ is a unit vector if –

   (i) λ = 1

   (ii) λ = −1

   (iii) a = 1/|λ|

   (iv) a = |λ|

   ›Reveal solutionSolution

   A unit vector has magnitude 1; set $|\lambda\vec a|=1$ and solve for $a=|\vec a|$.

   $\vec a$ has magnitude $a=|\vec a|$. Then $|\lambda\vec a| = |\lambda|\,|\vec a| = |\lambda|\,a$.

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set A11 markMCQ
   Q.The unit vector in the direction of $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ is

   (a) $\dfrac{\hat{i} - \hat{j} - 2\hat{k}}{6}$

   (b) $\dfrac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}}$

   (c) $\dfrac{\hat{i} - \hat{j} + 2\hat{k}}{6}$

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

   ›Reveal solutionSolution

   Divide $\vec a$ by its magnitude $\sqrt6$ to get the unit vector, so (b). …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2024Set ANNUAL1 mark
    Q.Find the value of x for which $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector.

    ›Reveal solutionSolution

    Set the magnitude of $x(\hat i+\hat j+\hat k)$ equal to 1 and solve for x.

    The vector is $x\hat i + x\hat j + x\hat k$. Its magnitude is:

    $\left|x(\hat i+\hat j+\hat k)\right| = \sqrt{x^2+x^2+x^2} = |x|\sqrt{3}$

    For this to be a unit vector: …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2024Set D1 markMCQ
    Q.If $\vec{a} = \vec{i} + \vec{j} + 2\vec{k}$, then the corresponding unit vector $\hat{a}$ in the direction of $\vec{a}$ is

    (a) $\frac{\vec{i} + \vec{j} + \vec{k}}{\sqrt{6}}$

    (b) $\frac{\vec{i} + \vec{j} + 2\vec{k}}{\sqrt{6}}$

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

    (d) $\frac{\vec{i} + \vec{j} + \vec{k}}{6}$

    ›Reveal solutionSolution

    Unit vector $= \vec{a}/|\vec{a}|$.

    $|\vec{a}| = |\vec{i} + \vec{j} + 2\vec{k}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}.$ Hence …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2024Set A1 mark
    Q.If $x\cdot(\hat{i}+\hat{j}+\hat{k})$ is a unit vector, write the value of $x$.

    ›Reveal solutionSolution

    If $x(\hat i+\hat j+\hat k)$ is a unit vector then $x=\pm\dfrac{1}{\sqrt3}$.

    The magnitude of $\hat i+\hat j+\hat k$ is $\sqrt{1^2+1^2+1^2}=\sqrt3$. For $x(\hat i+\hat j+\hat k)$ to be a unit vector we need $|x|\sqrt3=1$, …

    🔒 Start your 14-day free trial to see the full solution →