CBSE Class 12 Maths Exemplar Chapter 10: Vector Algebra NCERT Solutions

Let $\backslash vec\{c\}$ be the sum of the given vectors $\backslash vec\{a\}$ and $\backslash vec\{b\}$.

$$\backslash vec\{c\}\; =\; \backslash vec\{a\}\; +\; \backslash vec\{b\}\; =\; (2\backslash hat\{i\}\; -\; \backslash hat\{j\}\; +\; \backslash hat\{k\})\; +\; (2\backslash hat\{j\}\; +\; \backslash hat\{k\})$$

$$\backslash vec\{c\}\; =\; 2\backslash hat\{i\}\; +\; (-1+2)\backslash hat\{j\}\; +\; (1+1)\backslash hat\{k\}\; =\; 2\backslash hat\{i\}\; +\; \backslash hat\{j\}\; +\; 2\backslash hat\{k\}$$

To find the unit vector in the direction of $\backslash vec\{c\}$, we need to divide $\backslash vec\{c\}$ by its magnitude, $|\backslash vec\{c\}|$.

The magnitude of $\backslash vec\{c\}$ is calculated as:

$$|\backslash vec\{c\}|\; =\; \backslash sqrt\{2^2\; +\; 1^2\; +\; 2^2\}\; =\; \backslash sqrt\{4\; +\; 1\; +\; 4\}\; =\; \backslash sqrt\{9\}\; =\; 3$$

Therefore, the unit vector in the direction of $\backslash vec\{c\}$ is:

$$\backslash hat\{c\}\; =\; \backslash frac\{\backslash vec\{c\}\}\{|\backslash vec\{c\}|\}\; =\; \backslash frac\{2\backslash hat\{i\}\; +\; \backslash hat\{j\}\; +\; 2\backslash hat\{k\}\}\{3\}$$

Answer: The unit vector is $\backslash frac\{2\backslash hat\{i\}\; +\; \backslash hat\{j\}\; +\; 2\backslash hat\{k\}\}\{3\}$.

Given vectors are $\backslash vec\{a\}\; =\; \backslash hat\{i\}\; +\; \backslash hat\{j\}\; +\; 2\backslash hat\{k\}$ and $\backslash vec\{b\}\; =\; 2\backslash hat\{i\}\; +\; \backslash hat\{j\}\; -\; 2\backslash hat\{k\}$.

1. To find the unit vector in the direction of $\backslash vec\{b\}$, we first find the magnitude of $\backslash vec\{b\}$.

   $$|\backslash vec\{b\}|\; =\; \backslash sqrt\{2^2\; +\; 1^2\; +\; (-2)^2\}\; =\; \backslash sqrt\{4\; +\; 1\; +\; 4\}\; =\; \backslash sqrt\{9\}\; =\; 3$$

   The unit vector in the direction of $\backslash vec\{b\}$ is:

   $$\backslash hat\{b\}\; =\; \backslash frac\{\backslash vec\{b\}\}\{|\backslash vec\{b\}|\}\; =\; \backslash frac\{2\backslash hat\{i\}\; +\; \backslash hat\{j\}\; -\; 2\backslash hat\{k\}\}\{3\}$$

2. First, calculate the vector $2\backslash vec\{a\}\; -\; \backslash vec\{b\}$:

   $$2\backslash vec\{a\}\; -\; \backslash vec\{b\}\; =\; 2(\backslash hat\{i\}\; +\; \backslash hat\{j\}\; +\; 2\backslash hat\{k\})\; -\; (2\backslash hat\{i\}\; +\; \backslash hat\{j\}\; -\; 2\backslash hat\{k\})$$

   $$=\; (2\backslash hat\{i\}\; +\; 2\backslash hat\{j\}\; +\; 4\backslash hat\{k\})\; -\; (2\backslash hat\{i\}\; +\; \backslash hat\{j\}\; -\; 2\backslash hat\{k\})$$

   $$=\; (2-2)\backslash hat\{i\}\; +\; (2-1)\backslash hat\{j\}\; +\; (4-(-2))\backslash hat\{k\}\; =\; 0\backslash hat\{i\}\; +\; 1\backslash hat\{j\}\; +\; 6\backslash hat\{k\}\; =\; \backslash hat\{j\}\; +\; 6\backslash hat\{k\}$$

   Now, find the magnitude of this resultant vector:

   $$|2\backslash vec\{a\}\; -\; \backslash vec\{b\}|\; =\; \backslash sqrt\{0^2\; +\; 1^2\; +\; 6^2\}\; =\; \backslash sqrt\{0\; +\; 1\; +\; 36\}\; =\; \backslash sqrt\{37\}$$

   The unit vector in the direction of $2\backslash vec\{a\}\; -\; \backslash vec\{b\}$ is:

   $$\backslash frac\{2\backslash vec\{a\}\; -\; \backslash vec\{b\}\}\{|2\backslash vec\{a\}\; -\; \backslash vec\{b\}|\}\; =\; \backslash frac\{\backslash hat\{j\}\; +\; 6\backslash hat\{k\}\}\{\backslash sqrt\{37\}\}$$

Answer: (i) $\backslash frac\{2\backslash hat\{i\}\; +\; \backslash hat\{j\}\; -\; 2\backslash hat\{k\}\}\{3\}$, (ii) $\backslash frac\{\backslash hat\{j\}\; +\; 6\backslash hat\{k\}\}\{\backslash sqrt\{37\}\}$.

The coordinates of point P are (5, 0, 8) and the coordinates of point Q are (3, 3, 2).

The position vector of P is $\backslash vec\{OP\}\; =\; 5\backslash hat\{i\}\; +\; 0\backslash hat\{j\}\; +\; 8\backslash hat\{k\}$.

The position vector of Q is $\backslash vec\{OQ\}\; =\; 3\backslash hat\{i\}\; +\; 3\backslash hat\{j\}\; +\; 2\backslash hat\{k\}$.

The vector $\backslash vec\{PQ\}$ is given by the difference between the position vectors of Q and P:

$$\backslash vec\{PQ\}\; =\; \backslash vec\{OQ\}\; -\; \backslash vec\{OP\}\; =\; (3\backslash hat\{i\}\; +\; 3\backslash hat\{j\}\; +\; 2\backslash hat\{k\})\; -\; (5\backslash hat\{i\}\; +\; 0\backslash hat\{j\}\; +\; 8\backslash hat\{k\})$$

$$\backslash vec\{PQ\}\; =\; (3-5)\backslash hat\{i\}\; +\; (3-0)\backslash hat\{j\}\; +\; (2-8)\backslash hat\{k\}\; =\; -2\backslash hat\{i\}\; +\; 3\backslash hat\{j\}\; -\; 6\backslash hat\{k\}$$

To find the unit vector in the direction of $\backslash vec\{PQ\}$, we calculate its magnitude:

$$|\backslash vec\{PQ\}|\; =\; \backslash sqrt\{(-2)^2\; +\; 3^2\; +\; (-6)^2\}\; =\; \backslash sqrt\{4\; +\; 9\; +\; 36\}\; =\; \backslash sqrt\{49\}\; =\; 7$$

The unit vector in the direction of $\backslash vec\{PQ\}$ is:

$$\backslash frac\{\backslash vec\{PQ\}\}\{|\backslash vec\{PQ\}|\}\; =\; \backslash frac\{-2\backslash hat\{i\}\; +\; 3\backslash hat\{j\}\; -\; 6\backslash hat\{k\}\}\{7\}$$

Answer: The unit vector is $\backslash frac\{-2\backslash hat\{i\}\; +\; 3\backslash hat\{j\}\; -\; 6\backslash hat\{k\}\}\{7\}$.

Let the position vector of point A be $\backslash vec\{OA\}\; =\; \backslash vec\{a\}$ and the position vector of point B be $\backslash vec\{OB\}\; =\; \backslash vec\{b\}$.

The vector $\backslash vec\{BA\}$ is given by $\backslash vec\{OA\}\; -\; \backslash vec\{OB\}\; =\; \backslash vec\{a\}\; -\; \backslash vec\{b\}$.

We are given that point C lies on BA produced such that $BC\; =\; 1.5BA$. Since C is on the line BA produced, the vector $\backslash vec\{BC\}$ is in the same direction as $\backslash vec\{BA\}$.

Therefore, $$\backslash vec\{BC\}\; =\; 1.5\; \backslash vec\{BA\}\; =\; 1.5\; (\backslash vec\{a\}\; -\; \backslash vec\{b\})$$

Let the position vector of point C be $\backslash vec\{OC\}$. We know that $\backslash vec\{BC\}\; =\; \backslash vec\{OC\}\; -\; \backslash vec\{OB\}$.

Substituting the values, we get:

$$\backslash vec\{OC\}\; -\; \backslash vec\{b\}\; =\; 1.5\; (\backslash vec\{a\}\; -\; \backslash vec\{b\})$$

$$\backslash vec\{OC\}\; =\; \backslash vec\{b\}\; +\; 1.5\backslash vec\{a\}\; -\; 1.5\backslash vec\{b\}$$

$$\backslash vec\{OC\}\; =\; 1.5\backslash vec\{a\}\; +\; (1\; -\; 1.5)\backslash vec\{b\}$$

$$\backslash vec\{OC\}\; =\; 1.5\backslash vec\{a\}\; -\; 0.5\backslash vec\{b\}$$

This can also be written as:

$$\backslash vec\{OC\}\; =\; \backslash frac\{3\}\{2\}\backslash vec\{a\}\; -\; \backslash frac\{1\}\{2\}\backslash vec\{b\}\; =\; \backslash frac\{3\backslash vec\{a\}\; -\; \backslash vec\{b\}\}\{2\}$$

Answer: The position vector of point C is $1.5\backslash vec\{a\}\; -\; 0.5\backslash vec\{b\}$ or $\backslash frac\{3\backslash vec\{a\}\; -\; \backslash vec\{b\}\}\{2\}$.