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Q7:
Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.

Solution :

Step 1: Identification of the Vector

The given vector is $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$. In component form, this can be written as $\vec{a} = (1, 1, 2)$.

Step 2: Calculation of the Magnitude

The magnitude of a vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is defined by the formula: $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$ [By the definition of the Euclidean norm in $\mathbb{R}^3$].

Substituting the given components:

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

Step 3: Determining the Unit Vector

A unit vector $\hat{a}$ in the direction of vector $\vec{a}$ is defined as the vector divided by its own magnitude: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$ [Per the definition of normalized vectors].

Substituting the values derived in Step 1 and Step 2:

$\hat{a} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}}$ $\hat{a} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$

Geometric Representation

$\vec{a}$ x z y

Step 4: Verification (Optional)

To verify that $\hat{a}$ is indeed a unit vector, we calculate its magnitude: $|\hat{a}| = \sqrt{\left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{2}{\sqrt{6}}\right)^2}$ $|\hat{a}| = \sqrt{\frac{1}{6} + \frac{1}{6} + \frac{4}{6}} = \sqrt{\frac{6}{6}} = 1$ Since the magnitude is unity, the vector is correctly normalized.

Final Solution: The unit vector in the direction of $\vec{a}$ is $\frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$.


More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.2


CBSE Solutions for Class 12 Mathematics Vector Algebra


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