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Math Notes For Class 12 Chapter 7 Ex 7.1 Solved & Quiz
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Math Notes For Class 12 Chapter 7 Ex 7.1 Solved & Quiz
In 2nd-year mathematics (FSc Pre-Engineering), a vector in two or three dimensions is a quantity possessing both magnitude and direction. Vectors are fundamental in analytic geometry and physics for describing quantities like displacement, velocity, and force.
Key Definitions:
| Concept / Quantity | Mathematical Formula | Description |
|---|---|---|
| Vector Between Points $A(x_1, y_1)$ and $B(x_2, y_2)$ | $\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}$ | Displacement vector from initial point $A$ to terminal point $B$. |
| Magnitude of 2D Vector $\vec{v} = a\hat{i} + b\hat{j}$ | $|\vec{v}| = \sqrt{a^2 + b^2}$ | Euclidean length of the vector in the Cartesian plane. |
| Magnitude of 3D Vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ | $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$ | Euclidean distance from the origin in 3D space. |
| Unit Vector $\hat{v}$ | $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$ | Vector scaled to unit length pointing in the same direction. |
| Ratio Formula (Internal Division) | $\vec{r} = \frac{q\vec{a} + p\vec{b}}{p + q}$ | Position vector dividing segment $AB$ in the ratio $p:q$. |
Question 1 (Part i): Compute the magnitude of the vector $\vec{u} = 2\hat{i} - 7\hat{j}$.
Solution:
Given vector: $\vec{u} = 2\hat{i} - 7\hat{j}$
Comparing with $\vec{u} = a\hat{i} + b\hat{j}$, we have $a = 2$ and $b = -7$.
Using the magnitude formula $|\vec{u}| = \sqrt{a^2 + b^2}$:
$$|\vec{u}| = \sqrt{(2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53}$$
Answer: $$|\vec{u}| = \sqrt{53}$$
Question 1 (Part ii): Compute the magnitude of the vector $\vec{v} = \hat{i} + \hat{j}$.
Solution:
Given vector: $\vec{v} = 1\hat{i} + 1\hat{j}$
Here, $a = 1$ and $b = 1$.
$$|\vec{v}| = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$$
Answer: $$|\vec{v}| = \sqrt{2}$$
Question 1 (Part iii): Compute the magnitude of the 3D vector $\vec{w} = 3\hat{i} - 4\hat{j} + 12\hat{k}$.
Solution:
Given vector: $\vec{w} = 3\hat{i} - 4\hat{j} + 12\hat{k}$
Here, $a = 3$, $b = -4$, and $c = 12$.
$$|\vec{w}| = \sqrt{(3)^2 + (-4)^2 + (12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$$
Answer: $$|\vec{w}| = 13$$
Question 2 (Part i): Find the unit vector in the direction of $\vec{u} = \hat{i} + 2\hat{j}$.
Solution:
Given vector: $\vec{u} = \hat{i} + 2\hat{j}$
First, calculate the magnitude of $\vec{u}$:
$$|\vec{u}| = \sqrt{(1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5}$$
Now, calculate the unit vector $\hat{u} = \frac{\vec{u}}{|\vec{u}|}$:
$$\hat{u} = \frac{\hat{i} + 2\hat{j}}{\sqrt{5}} = \frac{1}{\sqrt{5}}\hat{i} + \frac{2}{\sqrt{5}}\hat{j}$$
Answer: $$\hat{u} = \frac{1}{\sqrt{5}}\hat{i} + \frac{2}{\sqrt{5}}\hat{j}$$
Question 2 (Part ii): Find the unit vector in the direction of $\vec{v} = 3\hat{i} - 4\hat{j}$.
Solution:
Given vector: $\vec{v} = 3\hat{i} - 4\hat{j}$
Calculate magnitude:
$$|\vec{v}| = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Apply the unit vector formula $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$:
$$\hat{v} = \frac{3\hat{i} - 4\hat{j}}{5} = \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}$$
Answer: $$\hat{v} = \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}$$
Question 2 (Part iii): Find the unit vector in the direction of $\vec{w} = -\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}$.
Solution:
Given vector: $\vec{w} = -\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}$
Calculate magnitude:
$$|\vec{w}| = \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{\frac{4}{4}} = 1$$
Since $|\vec{w}| = 1$, the given vector is already a unit vector:
$$\hat{w} = \frac{\vec{w}}{1} = -\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}$$
Answer: $$\hat{w} = -\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}$$
Question 3: Find the real values of $\alpha$ such that $|\alpha \hat{i} + (\alpha + 1)\hat{j} + 2\hat{k}| = 3$.
Solution:
Let $\vec{v} = \alpha \hat{i} + (\alpha + 1)\hat{j} + 2\hat{k}$.
Given condition: $|\vec{v}| = 3$.
Expressing the magnitude algebraically:
$$\sqrt{\alpha^2 + (\alpha + 1)^2 + (2)^2} = 3$$
Squaring both sides:
$$\alpha^2 + (\alpha^2 + 2\alpha + 1) + 4 = 9$$
$$2\alpha^2 + 2\alpha + 5 = 9$$
$$2\alpha^2 + 2\alpha - 4 = 0$$
Dividing the entire equation by $2$:
$$\alpha^2 + \alpha - 2 = 0$$
Factoring the quadratic equation:
$$(\alpha + 2)(\alpha - 1) = 0$$
Thus, $\alpha = 1$ or $\alpha = -2$.
Answer: $$\alpha = 1, -2$$
Question 4: Given points $A(1, -1)$, $B(4, 3)$, $C(-2, 5)$, and $D(2, 2)$, find the vector $\vec{AB} + \vec{CD}$ and its magnitude.
Solution:
First, calculate vector $\vec{AB}$:
$$\vec{AB} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} = (4 - 1)\hat{i} + (3 - (-1))\hat{j} = 3\hat{i} + 4\hat{j}$$
Next, calculate vector $\vec{CD}$:
$$\vec{CD} = (x_D - x_C)\hat{i} + (y_D - y_C)\hat{j} = (2 - (-2))\hat{i} + (2 - 5)\hat{j} = 4\hat{i} - 3\hat{j}$$
Now, compute the sum $\vec{u} = \vec{AB} + \vec{CD}$:
$$\vec{u} = (3\hat{i} + 4\hat{j}) + (4\hat{i} - 3\hat{j}) = (3 + 4)\hat{i} + (4 - 3)\hat{j} = 7\hat{i} + \hat{j}$$
Finally, compute the magnitude of $\vec{u}$:
$$|\vec{u}| = \sqrt{(7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$$
Answer: $$\vec{AB} + \vec{CD} = 7\hat{i} + \hat{j}, \quad |\vec{AB} + \vec{CD}| = 5\sqrt{2}$$
Question 5: Find the position vector of point $P$ which divides the line segment joining $A(2, 3)$ and $B(6, 7)$ internally in the ratio $2:3$.
Solution:
Let the position vectors of $A$ and $B$ be:
$$\vec{a} = 2\hat{i} + 3\hat{j}, \quad \vec{b} = 6\hat{i} + 7\hat{j}$$
Let the given ratio be $p:q = 2:3$, so $p = 2$ and $q = 3$.
Using the internal ratio formula $\vec{r} = \frac{q\vec{a} + p\vec{b}}{p + q}$:
$$\vec{r} = \frac{3(2\hat{i} + 3\hat{j}) + 2(6\hat{i} + 7\hat{j})}{2 + 3}$$
Expand the numerator:
$$\vec{r} = \frac{(6\hat{i} + 9\hat{j}) + (12\hat{i} + 14\hat{j})}{5}$$
Combine like components:
$$\vec{r} = \frac{18\hat{i} + 23\hat{j}}{5} = \frac{18}{5}\hat{i} + \frac{23}{5}\hat{j}$$
Answer: $$\vec{r} = \frac{18}{5}\hat{i} + \frac{23}{5}\hat{j}$$
Question 6 (Part i): Given vectors $\vec{u} = 2\hat{i} - \hat{j} + 3\hat{k}$, $\vec{v} = 3\hat{i} + 2\hat{j} - \hat{k}$, and $\vec{w} = \hat{i} + \hat{j} + \hat{k}$, find the linear combination $2\vec{u} - 3\vec{v} + 4\vec{w}$.
Solution:
Substitute the given vectors into the linear combination expression:
$$2\vec{u} - 3\vec{v} + 4\vec{w} = 2(2\hat{i} - \hat{j} + 3\hat{k}) - 3(3\hat{i} + 2\hat{j} - \hat{k}) + 4(\hat{i} + \hat{j} + \hat{k})$$
Expand each term:
$$= (4\hat{i} - 2\hat{j} + 6\hat{k}) - (9\hat{i} + 6\hat{j} - 3\hat{k}) + (4\hat{i} + 4\hat{j} + 4\hat{k})$$
Group components together:
$$= (4 - 9 + 4)\hat{i} + (-2 - 6 + 4)\hat{j} + (6 + 3 + 4)\hat{k}$$
$$= -\hat{i} - 4\hat{j} + 13\hat{k}$$
Answer: $$-\hat{i} - 4\hat{j} + 13\hat{k}$$
Question 6 (Part ii): For the same vectors $\vec{u} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{v} = 3\hat{i} + 2\hat{j} - \hat{k}$, find the unit vector in the direction of $2\vec{u} - \vec{v}$.
Solution:
First, calculate the vector $\vec{x} = 2\vec{u} - \vec{v}$:
$$\vec{x} = 2(2\hat{i} - \hat{j} + 3\hat{k}) - (3\hat{i} + 2\hat{j} - \hat{k})$$
$$\vec{x} = (4\hat{i} - 2\hat{j} + 6\hat{k}) - (3\hat{i} + 2\hat{j} - \hat{k})$$
$$\vec{x} = (4 - 3)\hat{i} + (-2 - 2)\hat{j} + (6 + 1)\hat{k} = \hat{i} - 4\hat{j} + 7\hat{k}$$
Next, find magnitude $|\vec{x}|$:
$$|\vec{x}| = \sqrt{(1)^2 + (-4)^2 + (7)^2} = \sqrt{1 + 16 + 49} = \sqrt{66}$$
Compute unit vector $\hat{x} = \frac{\vec{x}}{|\vec{x}|}$:
$$\hat{x} = \frac{\hat{i} - 4\hat{j} + 7\hat{k}}{\sqrt{66}} = \frac{1}{\sqrt{66}}\hat{i} - \frac{4}{\sqrt{66}}\hat{j} + \frac{7}{\sqrt{66}}\hat{k}$$
Answer: $$\hat{x} = \frac{1}{\sqrt{66}}\hat{i} - \frac{4}{\sqrt{66}}\hat{j} + \frac{7}{\sqrt{66}}\hat{k}$$
Q1: What is the magnitude of the vector $\vec{v} = 3\hat{i} - 4\hat{j}$?
Q2: If $\vec{a} = 2\hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - 3\hat{j}$, what is $\vec{a} - 2\vec{b}$?
Q3: What is the unit vector parallel to $\vec{u} = -\hat{i} + \hat{j}$?