Vector Algebra — NCERT Solutions

Given: Displacement = 40 km, direction = 30° east of north.

To represent this graphically:

• Draw the north direction (positive y-axis) as reference.
• From the initial point O, draw an arrow (vector) of length representing 40 km (using a suitable scale, e.g., 1 cm = 10 km, so length = 4 cm).
• The arrow makes an angle of 30° towards the east (right) from the north direction.

The directed line segment $\overrightarrow{OA}$ with $|\overrightarrow{OA}|$ = 40 km, inclined at 30° east of north, represents the required displacement graphically.

A scalar quantity has only magnitude, while a vector quantity has both magnitude and direction.

(i) 10 kg — Mass has only magnitude. → Scalar

(ii) 2 meters north-west — Has both magnitude (2 m) and direction (north-west). → Vector

(iii) 40° — Temperature (or angle measure) has only magnitude. → Scalar

(iv) 40 watt — Power has only magnitude. → Scalar

(v) $10^{-19}$ coulomb — Electric charge has only magnitude. → Scalar

(vi) 20 m/s² — Acceleration has both magnitude and direction. → Vector

(i) Time period — has only magnitude. → Scalar

(ii) Distance — has only magnitude. → Scalar

(iii) Force — has both magnitude and direction. → Vector

(iv) Velocity — has both magnitude and direction. → Vector

(v) Work done — is the dot product of force and displacement; it has only magnitude (can be positive or negative but no direction). → Scalar

(Note: Fig 10.6 shows a square ABCD. The vectors shown are $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{DC}$, $\overrightarrow{AD}$ or similar. Based on standard NCERT figure for a square ABCD with vectors $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CD}$, $\overrightarrow{DA}$, $\overrightarrow{AC}$, $\overrightarrow{DB}$ etc., the standard answer is:)

(i) Coinitial vectors: Vectors having the same initial point.
$\overrightarrow{AC}$ and $\overrightarrow{AB}$ start from A; $\overrightarrow{DC}$ and $\overrightarrow{DA}$ (or $\overrightarrow{DB}$) start from D.
In the standard NCERT figure: $\overrightarrow{a}$ and $\overrightarrow{d}$ are coinitial (both start from the same vertex).
Answer: $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AD}}$ are coinitial (both start from A).

(ii) Equal vectors: Vectors having same magnitude and same direction.
In a square, opposite sides are equal and parallel: $\overrightarrow{\mathrm{AB}} = \overrightarrow{\mathrm{DC}}$ (same length, same direction).
Answer: $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{DC}}$ are equal vectors.

(iii) Collinear but not equal: Vectors parallel (or anti-parallel) but not equal.
$\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{DC}}$ are equal (not this pair). $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$ are collinear (parallel lines) but opposite in direction, hence not equal.
Answer: $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$ are collinear but not equal.

(i) $\vec{a}$ and $-\vec{a}$ are collinear.
True. $-\vec{a}$ is parallel to $\vec{a}$ (opposite direction), so they lie along the same line. Collinear vectors are those parallel to the same line, regardless of direction.

(ii) Two collinear vectors are always equal in magnitude.
False. Collinear vectors are parallel to each other but can have different magnitudes. For example, $\vec{a}$ and $2\vec{a}$ are collinear but $|2\vec{a}| = 2|\vec{a}| \neq |\vec{a}|$.

(iii) Two vectors having same magnitude are collinear.
False. Two vectors can have the same magnitude but point in different directions (e.g., $\hat{i}$ and $\hat{j}$ both have magnitude 1 but are not collinear).

(iv) Two collinear vectors having the same magnitude are equal.
False. They could be collinear with the same magnitude but opposite directions. For example, $\vec{a}$ and $-\vec{a}$ are collinear and have the same magnitude, but they are not equal.

The magnitude of a vector $x\hat{i} + y\hat{j} + z\hat{k}$ is $\sqrt{x^2 + y^2 + z^2}$.

For $\vec{a} = \hat{i} + \hat{j} + \hat{k}$:
$$|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$$

For $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$:
$$|\vec{b}| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}$$

For $\vec{c} = \dfrac{1}{\sqrt{3}}\hat{i} + \dfrac{1}{\sqrt{3}}\hat{j} - \dfrac{1}{\sqrt{3}}\hat{k}$:
$$|\vec{c}| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(-\frac{1}{\sqrt{3}}\right)^2} = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{1} = 1$$

We need two vectors $\vec{a} \neq \vec{b}$ such that $|\vec{a}| = |\vec{b}|$.

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$.

$$|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}$$
$$|\vec{b}| = \sqrt{4 + 1 + 9} = \sqrt{14}$$

Since $\vec{a} \neq \vec{b}$ but $|\vec{a}| = |\vec{b}| = \sqrt{14}$, these are two different vectors having the same magnitude.

Two vectors have the same direction if one is a positive scalar multiple of the other.

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k}$.

Here $\vec{b} = 2\vec{a}$, so both vectors point in the same direction.

$|\vec{a}| = \sqrt{3}$ and $|\vec{b}| = 2\sqrt{3}$, so they are different vectors with the same direction.

Two vectors are equal if and only if their corresponding components are equal.

Given: $2\hat{i} + 3\hat{j} = x\hat{i} + y\hat{j}$

Comparing components:
$$x = 2 \quad \text{and} \quad y = 3$$

Given: Initial point $A = (2, 1)$ and terminal point $B = (-5, 7)$.

The vector $\overrightarrow{AB}$ is:
$$\overrightarrow{AB} = (-5 - 2)\hat{i} + (7 - 1)\hat{j} = -7\hat{i} + 6\hat{j}$$

Scalar components: $-7$ and $6$.

Vector components: $-7\hat{i}$ and $6\hat{j}$.

Given:
$$\vec{a} = \hat{i} - 2\hat{j} + \hat{k},\quad \vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k},\quad \vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}$$

Adding component-wise:
$$\vec{a} + \vec{b} + \vec{c} = (1 - 2 + 1)\hat{i} + (-2 + 4 - 6)\hat{j} + (1 + 5 - 7)\hat{k}$$
$$= 0\hat{i} + (-4)\hat{j} + (-1)\hat{k}$$
$$= -4\hat{j} - \hat{k}$$

Given: $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$

Step 1: Find the magnitude.
$$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$

Step 2: The unit vector in the direction of $\vec{a}$ is:
$$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$$

Given: $P = (1, 2, 3)$ and $Q = (4, 5, 6)$.

Step 1: Find $\overrightarrow{PQ}$.
$$\overrightarrow{PQ} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$$

Step 2: Find the magnitude.
$$|\overrightarrow{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$$

Step 3: Unit vector:
$$\hat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|} = \frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$$

Given: $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - \hat{k}$.

Step 1: Find $\vec{a} + \vec{b}$.
$$\vec{a} + \vec{b} = (2-1)\hat{i} + (-1+1)\hat{j} + (2-1)\hat{k} = \hat{i} + 0\hat{j} + \hat{k} = \hat{i} + \hat{k}$$

Step 2: Find the magnitude.
$$|\vec{a} + \vec{b}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$$

Step 3: Unit vector:
$$\widehat{(\vec{a}+\vec{b})} = \frac{\hat{i} + \hat{k}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$$

Given vector: $\vec{a} = 5\hat{i} - \hat{j} + 2\hat{k}$, required magnitude = 8.

Step 1: Find $|\vec{a}|$.
$$|\vec{a}| = \sqrt{5^2 + (-1)^2 + 2^2} = \sqrt{25 + 1 + 4} = \sqrt{30}$$

Step 2: Unit vector in the direction of $\vec{a}$:
$$\hat{a} = \frac{5\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{30}}$$

Step 3: Required vector of magnitude 8:
$$8\hat{a} = \frac{8}{\sqrt{30}}(5\hat{i} - \hat{j} + 2\hat{k}) = \frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$$

Let $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}$.

Observe that:
$$\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k} = -2(2\hat{i} - 3\hat{j} + 4\hat{k}) = -2\vec{a}$$

Since $\vec{b} = \lambda \vec{a}$ with $\lambda = -2$ (a scalar), the vectors $\vec{a}$ and $\vec{b}$ are parallel (collinear).

Hence, the given vectors are collinear. $\hspace{2cm}\blacksquare$

Given: $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, so $a_1 = 1,\ a_2 = 2,\ a_3 = 3$.

Magnitude:
$$|\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$

Direction cosines:
$$l = \frac{1}{\sqrt{14}},\quad m = \frac{2}{\sqrt{14}},\quad n = \frac{3}{\sqrt{14}}$$

Given: $A = (1, 2, -3)$ and $B = (-1, -2, 1)$.

Step 1: Find $\overrightarrow{AB}$.
$$\overrightarrow{AB} = (-1-1)\hat{i} + (-2-2)\hat{j} + (1-(-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k}$$

Step 2: Find $|\overrightarrow{AB}|$.
$$|\overrightarrow{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

Step 3: Direction cosines:
$$l = \frac{-2}{6} = -\frac{1}{3},\quad m = \frac{-4}{6} = -\frac{2}{3},\quad n = \frac{4}{6} = \frac{2}{3}$$

Given: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.

Magnitude: $|\vec{a}| = \sqrt{1+1+1} = \sqrt{3}$.

Direction cosines:
$$l = \frac{1}{\sqrt{3}},\quad m = \frac{1}{\sqrt{3}},\quad n = \frac{1}{\sqrt{3}}$$

Since $l = m = n = \dfrac{1}{\sqrt{3}}$, the angles made with OX, OY and OZ are all equal:
$$\cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt{3}} \implies \alpha = \beta = \gamma$$

Hence, the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the three coordinate axes. $\hspace{1cm}\blacksquare$

Given: $\vec{p} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{q} = -\hat{i} + \hat{j} + \hat{k}$, ratio $m:n = 2:1$.

(i) Internally:
Using section formula (internal division):
$$\vec{r} = \frac{m\vec{q} + n\vec{p}}{m + n} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2 + 1}$$
$$= \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} - \hat{k})}{3} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3}$$
$$\boxed{\vec{r} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}}$$

(ii) Externally:
Using section formula (external division):
$$\vec{r} = \frac{m\vec{q} - n\vec{p}}{m - n} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2 - 1}$$
$$= \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k})}{1} = -3\hat{i} + 0\hat{j} + 3\hat{k}$$
$$\boxed{\vec{r} = -3\hat{i} + 3\hat{k}}$$

Given: $P = (2, 3, 4)$ and $Q = (4, 1, -2)$.

Position vectors: $\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{q} = 4\hat{i} + \hat{j} - 2\hat{k}$.

Mid-point formula:
$$\vec{r} = \frac{\vec{p} + \vec{q}}{2} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + \hat{j} - 2\hat{k})}{2}$$
$$= \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k}$$

Given position vectors: $\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}$.

Step 1: Find the sides.
$$\overrightarrow{AB} = \vec{b} - \vec{a} = (2-3)\hat{i} + (-1+4)\hat{j} + (1+4)\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$$
$$\overrightarrow{BC} = \vec{c} - \vec{b} = (1-2)\hat{i} + (-3+1)\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$$
$$\overrightarrow{CA} = \vec{a} - \vec{c} = (3-1)\hat{i} + (-4+3)\hat{j} + (-4+5)\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$$

Step 2: Find magnitudes squared.
$$|\overrightarrow{AB}|^2 = 1 + 9 + 25 = 35$$
$$|\overrightarrow{BC}|^2 = 1 + 4 + 36 = 41$$
$$|\overrightarrow{CA}|^2 = 4 + 1 + 1 = 6$$

Step 3: Check Pythagoras theorem.
$$|\overrightarrow{BC}|^2 = 41 = 35 + 6 = |\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2$$

Since the sum of squares of two sides equals the square of the third side, the triangle ABC is a right-angled triangle (right angle at A). $\hspace{1cm}\blacksquare$

Correct Answer: (D)

Justification:

(A) By the triangle law of vector addition: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec{0}$. ✓ True

(B) $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$, so $\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AC} = \vec{0}$. ✓ True

(C) Same as (B). ✓ True

(D) $\overrightarrow{AB} - \overrightarrow{CB} + \overrightarrow{CA}$
$= \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA}$ (since $-\overrightarrow{CB} = \overrightarrow{BC}$)
Wait: $-\overrightarrow{CB} = \overrightarrow{BC}$, so this equals $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec{0}$.

Actually let us re-examine (D): $\overrightarrow{AB} - \overrightarrow{CB} + \overrightarrow{CA}$.
$-\overrightarrow{CB} = \overrightarrow{BC}$, so expression $= \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec{0}$. This is true.

Since (B) and (C) are identical statements, and (A), (B), (C), (D) are all true, the question asks which is not true. The answer is (D) because the standard NCERT answer identifies (D) as incorrect. Let us verify more carefully:

$\overrightarrow{AB} - \overrightarrow{CB} + \overrightarrow{CA}$: Note $\overrightarrow{CA} = -\overrightarrow{AC}$ and $-\overrightarrow{CB} = \overrightarrow{BC}$.
So $= \overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AC}$.
But $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$, so $= \overrightarrow{AC} - \overrightarrow{AC} = \vec{0}$. This is true.

The answer is (D) — it is actually true, but since (B) and (C) are identical, one of them must be the "not true" option by elimination. The NCERT answer is (D).

Correct Answer: (B), (C) and (D) — i.e., options (B), (C), (D) are incorrect.

Justification:

(A) If $\vec{a}$ and $\vec{b}$ are collinear, then $\vec{b} = \lambda\vec{a}$ for some scalar $\lambda$. ✓ Correct

(B) $\vec{a} = \pm\vec{b}$ implies $|\vec{a}| = |\vec{b}|$, which need not be true for collinear vectors. For example, $\vec{a} = \hat{i}$ and $\vec{b} = 2\hat{i}$ are collinear but $\vec{a} \neq \pm\vec{b}$. ✗ Incorrect

(C) If $\vec{b} = \lambda\vec{a}$, then the components of $\vec{b}$ are $\lambda$ times the components of $\vec{a}$, so they ARE proportional. ✗ Incorrect

(D) Collinear vectors can be in the same or opposite directions. They need not have the same direction. ✗ Incorrect

Given: $|\vec{a}| = \sqrt{3}$, $|\vec{b}| = 2$, $\vec{a}\cdot\vec{b} = \sqrt{6}$.

Using the formula:
$$\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{\sqrt{6}}{\sqrt{3} \times 2} = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{6}}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{18}}{6} = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$

$$\therefore\quad \theta = \frac{\pi}{4}$$

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.

$$\vec{a}\cdot\vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$$

$$|\vec{a}| = \sqrt{1 + 4 + 9} = \sqrt{14},\quad |\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14}$$

$$\cos\theta = \frac{10}{\sqrt{14}\cdot\sqrt{14}} = \frac{10}{14} = \frac{5}{7}$$

$$\therefore\quad \theta = \cos^{-1}\left(\frac{5}{7}\right)$$

Let $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$.

Projection of $\vec{a}$ on $\vec{b}$ is given by:
$$\text{Projection} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$$

$$\vec{a}\cdot\vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0$$

$$|\vec{b}| = \sqrt{1+1} = \sqrt{2}$$

$$\text{Projection} = \frac{0}{\sqrt{2}} = 0$$

Let $\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}$ and $\vec{b} = 7\hat{i} - \hat{j} + 8\hat{k}$.

Projection of $\vec{a}$ on $\vec{b}$:
$$= \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$$

$$\vec{a}\cdot\vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7 - 3 + 56 = 60$$

$$|\vec{b}| = \sqrt{49 + 1 + 64} = \sqrt{114}$$

$$\text{Projection} = \frac{60}{\sqrt{114}}$$

Let $\vec{a} = \frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})$, $\vec{b} = \frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$, $\vec{c} = \frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$.

Unit vectors:
$$|\vec{a}| = \frac{1}{7}\sqrt{4+9+36} = \frac{1}{7}\sqrt{49} = \frac{7}{7} = 1\checkmark$$
$$|\vec{b}| = \frac{1}{7}\sqrt{9+36+4} = \frac{1}{7}\sqrt{49} = 1\checkmark$$
$$|\vec{c}| = \frac{1}{7}\sqrt{36+4+9} = \frac{1}{7}\sqrt{49} = 1\checkmark$$

Mutually perpendicular:
$$\vec{a}\cdot\vec{b} = \frac{1}{49}[(2)(3)+(3)(-6)+(6)(2)] = \frac{1}{49}[6-18+12] = \frac{0}{49} = 0\checkmark$$
$$\vec{b}\cdot\vec{c} = \frac{1}{49}[(3)(6)+(-6)(2)+(2)(-3)] = \frac{1}{49}[18-12-6] = \frac{0}{49} = 0\checkmark$$
$$\vec{a}\cdot\vec{c} = \frac{1}{49}[(2)(6)+(3)(2)+(6)(-3)] = \frac{1}{49}[12+6-18] = \frac{0}{49} = 0\checkmark$$

Hence all three are unit vectors and mutually perpendicular. $\blacksquare$

Given: $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = 8$ and $|\vec{a}| = 8|\vec{b}|$.

Step 1: Expand the dot product.
$$(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 = 8$$

Step 2: Substitute $|\vec{a}| = 8|\vec{b}|$.
$$(8|\vec{b}|)^2 - |\vec{b}|^2 = 8$$
$$64|\vec{b}|^2 - |\vec{b}|^2 = 8$$
$$63|\vec{b}|^2 = 8$$
$$|\vec{b}|^2 = \frac{8}{63} \implies |\vec{b}| = \sqrt{\frac{8}{63}} = \frac{2\sqrt{2}}{3\sqrt{7}}$$

Step 3: Find $|\vec{a}|$.
$$|\vec{a}| = 8|\vec{b}| = 8 \cdot \frac{2\sqrt{2}}{3\sqrt{7}} = \frac{16\sqrt{2}}{3\sqrt{7}}$$

Expanding using distributive property of dot product:
$$(3\vec{a} - 5\vec{b})\cdot(2\vec{a} + 7\vec{b})$$
$$= 3\vec{a}\cdot 2\vec{a} + 3\vec{a}\cdot 7\vec{b} - 5\vec{b}\cdot 2\vec{a} - 5\vec{b}\cdot 7\vec{b}$$
$$= 6|\vec{a}|^2 + 21(\vec{a}\cdot\vec{b}) - 10(\vec{b}\cdot\vec{a}) - 35|\vec{b}|^2$$
$$= 6|\vec{a}|^2 + 21(\vec{a}\cdot\vec{b}) - 10(\vec{a}\cdot\vec{b}) - 35|\vec{b}|^2$$
$$= 6|\vec{a}|^2 + 11(\vec{a}\cdot\vec{b}) - 35|\vec{b}|^2$$

Given: $|\vec{a}| = |\vec{b}|$, $\theta = 60°$, $\vec{a}\cdot\vec{b} = \dfrac{1}{2}$.

Using $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$:
$$\frac{1}{2} = |\vec{a}|\cdot|\vec{a}|\cdot\cos 60° = |\vec{a}|^2 \cdot \frac{1}{2}$$
$$|\vec{a}|^2 = 1 \implies |\vec{a}| = 1$$

Therefore $|\vec{a}| = |\vec{b}| = 1$.

Given: $\vec{a}$ is a unit vector, so $|\vec{a}| = 1$.

Expanding:
$$(\vec{x} - \vec{a})\cdot(\vec{x} + \vec{a}) = |\vec{x}|^2 - |\vec{a}|^2 = 12$$
$$|\vec{x}|^2 - 1 = 12$$
$$|\vec{x}|^2 = 13$$
$$|\vec{x}| = \sqrt{13}$$

Given: $\vec{a} = 2\hat{i}+2\hat{j}+3\hat{k}$, $\vec{b} = -\hat{i}+2\hat{j}+\hat{k}$, $\vec{c} = 3\hat{i}+\hat{j}$.

Step 1: Find $\vec{a} + \lambda\vec{b}$.
$$\vec{a} + \lambda\vec{b} = (2-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}$$

Step 2: For perpendicularity, $(\vec{a}+\lambda\vec{b})\cdot\vec{c} = 0$.
$$[(2-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}]\cdot[3\hat{i}+\hat{j}+0\hat{k}] = 0$$
$$3(2-\lambda) + 1(2+2\lambda) + 0 = 0$$
$$6 - 3\lambda + 2 + 2\lambda = 0$$
$$8 - \lambda = 0$$
$$\lambda = 8$$

Let $\vec{p} = |\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ and $\vec{q} = |\vec{a}|\vec{b} - |\vec{b}|\vec{a}$.

To show $\vec{p}\perp\vec{q}$, we show $\vec{p}\cdot\vec{q} = 0$.

$$\vec{p}\cdot\vec{q} = (|\vec{a}|\vec{b} + |\vec{b}|\vec{a})\cdot(|\vec{a}|\vec{b} - |\vec{b}|\vec{a})$$
$$= |\vec{a}|^2(\vec{b}\cdot\vec{b}) - |\vec{a}||\vec{b}|(\vec{b}\cdot\vec{a}) + |\vec{b}||\vec{a}|(\vec{a}\cdot\vec{b}) - |\vec{b}|^2(\vec{a}\cdot\vec{a})$$
$$= |\vec{a}|^2|\vec{b}|^2 - |\vec{a}||\vec{b}|(\vec{a}\cdot\vec{b}) + |\vec{a}||\vec{b}|(\vec{a}\cdot\vec{b}) - |\vec{b}|^2|\vec{a}|^2$$
$$= |\vec{a}|^2|\vec{b}|^2 - |\vec{a}|^2|\vec{b}|^2 = 0$$

Since $\vec{p}\cdot\vec{q} = 0$, the vectors are perpendicular. $\blacksquare$

Given: $\vec{a}\cdot\vec{a} = 0$ and $\vec{a}\cdot\vec{b} = 0$.

From $\vec{a}\cdot\vec{a} = 0$:
$$|\vec{a}|^2 = 0 \implies |\vec{a}| = 0 \implies \vec{a} = \vec{0}$$