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Q9:
Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
Solution :
Step 1: Coordinate Representation and Vector Formation
Given the vertices of the triangle in $\mathbb{R}^3$: $A = (1, 1, 2)$, $B = (2, 3, 5)$, and $C = (1, 5, 5)$.
To determine the area of triangle $ABC$, we represent two sides originating from a common vertex as vectors. We select vertex $A$ as the reference point:
$\vec{AB} = (2 - 1)\hat{i} + (3 - 1)\hat{j} + (5 - 2)\hat{k} = 1\hat{i} + 2\hat{j} + 3\hat{k}$
$\vec{AC} = (1 - 1)\hat{i} + (5 - 1)\hat{j} + (5 - 2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}$
Step 2: Theoretical Foundation for Area Calculation
[Per the definition of the cross product in vector algebra], the area of a triangle defined by two vectors $\vec{AB}$ and $\vec{AC}$ is given by the formula:
$\text{Area} = \frac{1}{2} | \vec{AB} \times \vec{AC} |$
Step 3: Computation of the Cross Product
The cross product $\vec{AB} \times \vec{AC}$ is calculated using the determinant form:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix}$
Expanding along the first row: $= \hat{i}((2)(3) - (3)(4)) - \hat{j}((1)(3) - (3)(0)) + \hat{k}((1)(4) - (2)(0))$ $= \hat{i}(6 - 12) - \hat{j}(3 - 0) + \hat{k}(4 - 0)$ $= -6\hat{i} - 3\hat{j} + 4\hat{k}$
Step 4: Magnitude Calculation
The magnitude of the resulting vector $| \vec{AB} \times \vec{AC} |$ is: $| -6\hat{i} - 3\hat{j} + 4\hat{k} | = \sqrt{(-6)^2 + (-3)^2 + (4)^2}$ $= \sqrt{36 + 9 + 16} = \sqrt{61}$
Step 5: Final Area Determination
Substituting the magnitude back into the area formula: $\text{Area} = \frac{1}{2} \sqrt{61}$ square units.
Final Solution: The area of the triangle is $\frac{\sqrt{61}}{2}$ square units.
More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.4
- Q1: Find $|\vec{a} \times \vec{b}|$, if $\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$.
- Q2: Find a unit vector perpendicular to each of the vector $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.
- Q3: If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$, $\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, then find $\theta$ and hence, the components of $\vec{a}$.
- Q4: Show that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$
- Q5: Find $\lambda$ and $\mu$ if $(2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}$.
- Q6: Given that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?
- Q7: Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, $c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Then show that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.
- Q8: If either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \times \vec{b} = \vec{0}$. Is the converse true? Justify your answer with an example.
- Q10: Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.
- Q11: Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is
- Q12: Area of a rectangle having vertices A, B, C and D with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$, respectively is
CBSE Solutions for Class 12 Mathematics Vector Algebra
Chapters in CBSE - Class 12 Mathematics
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