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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
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
- Q1(i): Compute the magnitude of the following vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$
- Q1(ii): Compute the magnitude of the following vectors: $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$
- Q1(iii): Compute the magnitude of the following vectors: $\vec{c} = \frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{1}{3}\hat{k}$
- Q2: Write two different vectors having same magnitude.
- Q3: Write two different vectors having same direction.
- Q4: Find the values of $x$ and $y$ so that the vectors $2\hat{i} + 3\hat{j}$ and $x\hat{i} + y\hat{j}$ are equal.
- Q5: Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).
- Q6: Find the sum of the vectors $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}$.
- Q8: Find the unit vector in the direction of vector $\vec{PQ}$, where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
- Q9: For given vectors, $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - \hat{k}$, find the unit vector in the direction of the vector $\vec{a} + \vec{b}$.
- Q10: Find a vector in the direction of vector $5\hat{i} - \hat{j} + 2\hat{k}$ which has magnitude 8 units.
- Q11: Show that the vectors $2\hat{i} - 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} - 8\hat{k}$ are collinear.
- Q12: Find the direction cosines of the vector $\hat{i} + 2\hat{j} + 3\hat{k}$.
- Q13: Find the direction cosines of the vector joining the points A (1, 2, –3) and B (–1, –2, 1), directed from A to B.
- Q14: Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes OX, OY and OZ.
- Q15(i): Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i} + 2\hat{j} - \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio 2 : 1 (i) internally
- Q15(ii): Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i} + 2\hat{j} - \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio 2 : 1 (ii) externally
- Q16: Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).
- Q17: Show that the points A, B and C with position vectors, $\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}$, respectively form the vertices of a right angled triangle.
- Q18: In triangle ABC (Fig 10.18), which of the following is not true:
- Q19: If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:
CBSE Solutions for Class 12 Mathematics Vector Algebra
Chapters in CBSE - Class 12 Mathematics
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