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Q14:
Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes OX, OY and OZ.
Solution :
Step 1: Define the Given Vector
Let the given vector be $\vec{a}$. We define the components as follows:
$\vec{a} = \hat{i} + \hat{j} + \hat{k}$
The components of the vector along the $x, y,$ and $z$ axes are $a_x = 1, a_y = 1,$ and $a_z = 1$, respectively. [By Definition of Cartesian Components of a Vector].
Step 2: Calculate the Magnitude of the Vector
The magnitude of the vector $\vec{a}$, denoted as $|\vec{a}|$, is calculated using the Euclidean norm formula:
$|\vec{a}| = \sqrt{(a_x)^2 + (a_y)^2 + (a_z)^2}$
$|\vec{a}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$
Step 3: Determine the Direction Cosines
If a vector makes angles $\alpha, \beta, \gamma$ with the $OX, OY,$ and $OZ$ axes, its direction cosines are given by $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$. These are defined as:
$\cos \alpha = \frac{a_x}{|\vec{a}|} = \frac{1}{\sqrt{3}}$
$\cos \beta = \frac{a_y}{|\vec{a}|} = \frac{1}{\sqrt{3}}$
$\cos \gamma = \frac{a_z}{|\vec{a}|} = \frac{1}{\sqrt{3}}$
[By Definition of Direction Cosines for a Vector in 3D Space].
Step 4: Analyze Equality of Inclination
Since $\cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}}$, and considering the domain of angles $\alpha, \beta, \gamma \in [0, \pi]$, the values of the angles must be equal:
$\alpha = \beta = \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Because the angles made with the coordinate axes are identical, the vector is equally inclined to the axes $OX, OY,$ and $OZ$.
Visual Representation
Final Solution: Since the direction cosines $\cos \alpha, \cos \beta, \cos \gamma$ are all equal to $\frac{1}{\sqrt{3}}$, the angles $\alpha, \beta, \gamma$ are equal; thus, the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.
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 $\vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}$.
- Q7: Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.
- 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.
- 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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