# Scalar Product

If $\overrightarrow{p}$.$\overrightarrow{q}$ = 6, $\vert \overrightarrow{q} \vert$ = 2$\sqrt 3$ unit and the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$ is 30°, find the length of $\overrightarrow{p}$.

Given,
$\vert \overrightarrow{q} \vert$ = 2$\sqrt 3$
$\overrightarrow{p}$.$\overrightarrow{q}$ = 6
$\vert \overrightarrow{p} \vert$ = ?

Let, θ = 30° be the angle between given vector. Then from cosine angle, cosθ = $\frac{\overrightarrow{p} . \overrightarrow{q}}{\vert \overrightarrow{p} \vert \vert \overrightarrow{q} \vert}$ or, cos30° = $\frac{6}{\vert \overrightarrow{p} \vert 2\sqrt 3}$
or, $\vert \overrightarrow{p} \vert$ = $\frac{6}{\frac{\sqrt 3}{2} \times 2\sqrt 3}$
$\vert \overrightarrow{p} \vert$ = 2 unit

If $\vert \overrightarrow{a} \vert$ = 4, $\vert \overrightarrow{b} \vert$ = 5 and $\overrightarrow{a} . \overrightarrow{b}$ = 10, then find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.

Given,
$\vert \overrightarrow{a} \vert$ = 4
$\vert \overrightarrow{b} \vert$ = 5
$\overrightarrow{a} . \overrightarrow{b}$ = 10

Let, θ be the angle between given vector. Then from cosine angle,
cosθ = $\frac{\overrightarrow{a} . \overrightarrow{b}}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ = $\frac{10}{20}$ = cos60°

∴ θ = 60°

If $\vert \overrightarrow{a} \vert$ = 3$\sqrt 3$, $\vert \overrightarrow{b} \vert$ = 4 and angle between $\vert \overrightarrow{a} \vert$ and $\vert \overrightarrow{b} \vert$, θ = 60° then find the value of $\overrightarrow{a}.\overrightarrow{b}$.

Given,
$\vert \overrightarrow{a} \vert$ = 3$\sqrt 3$
$\vert \overrightarrow{a} \vert$ = 4
Angle between $\vert \overrightarrow{a} \vert$ amd $\vert \overrightarrow{b} \vert$, θ = 60°

Now, from cosine angle between two vectors, cosθ = $\frac{\overrightarrow{a} . \overrightarrow{b}}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ or, cos60° = $\frac{\overrightarrow{a} . \overrightarrow{b}}{12\sqrt{3}}$
or, $\frac{12\sqrt{3}}{2}$ = $\overrightarrow{a} . \overrightarrow{b}$
$\overrightarrow{a} . \overrightarrow{b}$ = 6$\sqrt3$

10$\overrightarrow{i}$ - 7$\overrightarrow{j}$ and a$\overrightarrow{i}$ + 10$\overrightarrow{j}$ are perpendicular to each other, find the value of a.

Let,
$\overrightarrow{a}$ = 10$\overrightarrow{i}$ - 7$\overrightarrow{j}$
$\overrightarrow{b}$ = a$\overrightarrow{i}$ + 10$\overrightarrow{j}$

Since, $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$, $\overrightarrow{a} . \overrightarrow{b}$ = 0 or, (10$\overrightarrow{i}$ - 7$\overrightarrow{j}$).(a$\overrightarrow{i}$ + 10$\overrightarrow{j}$) = 0
or, 10a($\overrightarrow{i}$)2 + 100($\overrightarrow{i}.\overrightarrow{j}$) - 7a($\overrightarrow{j}.\overrightarrow{i}$) - 70($\overrightarrow{j}$)2 = 0
We know, $\overrightarrow{i}$.$\overrightarrow{i}$ = 1, $\overrightarrow{j}$.$\overrightarrow{j}$ = 1, $\overrightarrow{i}$.$\overrightarrow{j}$ = $\overrightarrow{j}$.$\overrightarrow{i}$ = 0
so, 10a + 0 - 0 - 70 = 0
∴ a = 7.

If $\overrightarrow{a}$ = 4$\overrightarrow{i}$ + 2$\overrightarrow{j}$ and $\overrightarrow{b}$ = -$\overrightarrow{i}$ + 2$\overrightarrow{j}$, then find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.

Here,
$\overrightarrow{a}$ = 4$\overrightarrow{i}$ + 2$\overrightarrow{j}$
$\overrightarrow{b}$ = -$\overrightarrow{i}$ + 2$\overrightarrow{j}$

Now,
$\overrightarrow{a} . \overrightarrow{b}$ = (4$\overrightarrow{i}$ + 2$\overrightarrow{j}$) . (-$\overrightarrow{i}$ + 2$\overrightarrow{j}$) = -4 + 4 = 0

Let, θ be the angle between given vector. Then from cosine angle,
cosθ = $\frac{\overrightarrow{a} . \overrightarrow{b}}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ = $\frac{0}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ = cos90°

∴ θ = 90°

If $\vert \overrightarrow{b} \vert$ = 6, $\overrightarrow{a}.\overrightarrow{b}$ = 12 and angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is 60°, find the value of $\vert \overrightarrow{a} \vert$ .

Given,
$\vert \overrightarrow{b} \vert$ = 6
$\overrightarrow{a}.\overrightarrow{b}$ = 12
Angle between $\overrightarrow{a}$ and $\overrightarrow{b}$(θ) = 60°

Now, from cosine angle between two vectors, cosθ = $\frac{\overrightarrow{a} . \overrightarrow{b}}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ or, cos60° = $\frac{12}{6\vert \overrightarrow{a} \vert}$
$\overrightarrow{a}$ = 2 × 2 = 4

If $\overrightarrow{p}$.$\overrightarrow{q}$ = 18$\sqrt 3$, $\vert \overrightarrow{p} \vert$ = 6 and $\vert \overrightarrow{q} \vert$ = 6, find the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$.

Given,
$\vert \overrightarrow{p} \vert$ = 6
$\vert \overrightarrow{q} \vert$ = 6
$\overrightarrow{p} . \overrightarrow{q}$ = 18$\sqrt 3$

Let, θ be the angle between given vector. Then from cosine angle,
cosθ = $\frac{\overrightarrow{a} . \overrightarrow{b}}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ = $\frac{18\sqrt 3}{36}$ = $\frac{\sqrt 3}{2}$ = cos30°

∴ θ = 30°

If $\overrightarrow{p}$ + $\overrightarrow{q}$ + $\overrightarrow{r}$ = 0, $\vert \overrightarrow{p} \vert$ = 6, $\vert \overrightarrow{q} \vert$ = 7 and $\vert \overrightarrow{r} \vert$ = $\sqrt {127}$, find the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$.

Given,
$\vert \overrightarrow{p} \vert$ = 6
$\vert \overrightarrow{q} \vert$ = 7
$\vert \overrightarrow{r} \vert$ = $\sqrt {127}$

Now, $\overrightarrow{p}$ + $\overrightarrow{q}$ + $\overrightarrow{r}$ = 0 or, $\overrightarrow{p}$ + $\overrightarrow{q}$ = -$\overrightarrow{r}$
Multiplying by same vector on both sides,
or, $\vert \overrightarrow{p} + \overrightarrow{q} \vert$2 = $\vert -\overrightarrow{r} \vert$2
or, $\vert \overrightarrow{p} \vert$2 + 2$\overrightarrow{p}.\overrightarrow{q}$ + $\vert \overrightarrow{q} \vert$2 = $\vert \overrightarrow{r} \vert$2
or, 36 + 2$\overrightarrow{p}.\overrightarrow{q}$ + 49 = 127
so, $\overrightarrow{p}.\overrightarrow{q}$ = 21

Let, θ be the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$. Then from cosine angle,
cosθ = $\frac{\overrightarrow{a} . \overrightarrow{b}}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ = $\frac{21}{42}$ = $\frac 12$ = cos60°

∴ θ = 60°

If $\overrightarrow{a}$ = -5$\overrightarrow{i}$ + 3$\overrightarrow{j}$ and $\overrightarrow{b}$ = p$\overrightarrow{i}$ + (p + 2)$\overrightarrow{j}$ are perpendicular to each other, find the value of p.

Given,
$\overrightarrow{a}$ = -5$\overrightarrow{i}$ + 3$\overrightarrow{j}$
$\overrightarrow{b}$ = p$\overrightarrow{i}$ + (p + 2)$\overrightarrow{j}$

Since, $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$, $\overrightarrow{a}$.$\overrightarrow{b}$ = 0 or, (-5$\overrightarrow{i}$ + 3$\overrightarrow{j}$).{p$\overrightarrow{i}$ + (p + 2)$\overrightarrow{j}$} = 0
or, -5p($\overrightarrow{i}$)2 - 5(p + 2)($\overrightarrow{i}$.$\overrightarrow{j}$) + 3p($\overrightarrow{j}$.$\overrightarrow{i}$) + 3(p + 2)($\overrightarrow{j}$)2 = 0
We know, $\overrightarrow{i}$.$\overrightarrow{i}$ = 1, $\overrightarrow{j}$.$\overrightarrow{j}$ = 1, $\overrightarrow{i}$.$\overrightarrow{j}$ = $\overrightarrow{j}$.$\overrightarrow{i}$ = 0
or, -5p + 3(p + 2) = 0
or, 3p + 6 = 5p
so, p = 3

Find the angle between unit vector $\overrightarrow{i}$ and $\overrightarrow{a}$ = $\sqrt{3} \overrightarrow{i}$ + $\overrightarrow{j}$.

Let, θ be the angle between unit vector $\overrightarrow{i}$ and $\overrightarrow{a}$ = $\sqrt{3} \overrightarrow{i}$ + $\overrightarrow{j}$.

From cosine angle,
cosθ = $\frac{\overrightarrow{i}.\overrightarrow{a}}{\vert \overrightarrow{i} \vert \vert \overrightarrow{a} \vert}$ = $\frac{\overrightarrow{i}.(\sqrt3 \overrightarrow{i} + \overrightarrow{j})}{1\sqrt{(\sqrt{3})^2 + 1^2}}$
= $\frac{\sqrt{3}(\overrightarrow{i})^2 + \overrightarrow{i}.\overrightarrow{j}}{2}$
= $\frac{\sqrt3}{2}$

So, θ = 30°

If $\overrightarrow{a}$ + 2$\overrightarrow{b}$ and 5$\overrightarrow{a}$ - 4$\overrightarrow{b}$ are perpendicular to each other and $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors, find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.

Let,
$\overrightarrow{p}$ = $\overrightarrow{a}$ + 2$\overrightarrow{b}$
$\overrightarrow{q}$ = 5$\overrightarrow{a}$ - 4$\overrightarrow{b}$

Since, $\overrightarrow{p}$ is perpendicular to $\overrightarrow{q}$, $\overrightarrow{p}$.$\overrightarrow{q}$ = 0 or, ($\overrightarrow{a}$ + 2$\overrightarrow{b}$).(5$\overrightarrow{a}$ - 4$\overrightarrow{b}$) = 0
or, 5($\overrightarrow{a}$)2 - 4($\overrightarrow{a}$.$\overrightarrow{b}$) + 10($\overrightarrow{b}$.$\overrightarrow{a}$) - 8($\overrightarrow{b}$)2 = 0
$\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vector. so, ($\overrightarrow{a}$)2 = ($\overrightarrow{b}$)2 = 1
or, -3 + 6$\overrightarrow{a}$.$\overrightarrow{b}$ = 0
so, $\overrightarrow{a}$.$\overrightarrow{b}$ = $\frac 12$

Let, θ be the angle between given vector. Then from cosine angle,
cosθ = $\frac{\overrightarrow{a} . \overrightarrow{b}}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ = $\frac{\frac 12}{1}$ = cos60°

so, θ = 60°

If $\vert \overrightarrow{a} \vert$ = 5$\sqrt3$, $\vert \overrightarrow{b} \vert$ = 6 and θ = 30°, find the value of $\overrightarrow{a}.\overrightarrow{b}$.

Given,
$\vert \overrightarrow{a} \vert$ = 5$\sqrt3$
$\vert \overrightarrow{b} \vert$ = 6
θ = 30°

From cosine angle, cosθ = $\frac{\overrightarrow{a} . \overrightarrow{b}}{\vert \overrightarrow{a} \vert \vert \overrightarrow{b} \vert}$ or, cos30° = $\frac{\overrightarrow{a} . \overrightarrow{b}}{30\sqrt{3}}$
or, $\frac{\sqrt3}{2}$ × 30$\sqrt3$ = $\overrightarrow{a}$ . $\overrightarrow{b}$
$\overrightarrow{a}$ . $\overrightarrow{b}$ = 45

If $\overrightarrow{p}$ = 2$\overrightarrow{i}$ + 3$\overrightarrow{j}$, $\overrightarrow{q}$ = -a$\overrightarrow{i}$ + 4$\overrightarrow{j}$ and $\overrightarrow{p}.\overrightarrow{q}$ = 0, find the value of a.

Given, $\overrightarrow{p}$ = 2$\overrightarrow{i}$ + 3$\overrightarrow{j}$
$\overrightarrow{q}$ = -a$\overrightarrow{i}$ + 4$\overrightarrow{j}$
$\overrightarrow{p}.\overrightarrow{q}$ = 0

Now, $\overrightarrow{p}.\overrightarrow{q}$ = 0 or, (2$\overrightarrow{i}$ + 3$\overrightarrow{j}$).(-a$\overrightarrow{i}$ + 4$\overrightarrow{j}$) = 0
or, -2a($\overrightarrow{i}$)2 + 8($\overrightarrow{i}$.$\overrightarrow{j}$) - 3a($\overrightarrow{j}$.$\overrightarrow{i}$) + 12($\overrightarrow{j}$)2 = 0
We know, $\overrightarrow{i}$.$\overrightarrow{i}$ = 1, $\overrightarrow{j}$.$\overrightarrow{j}$ = 1, $\overrightarrow{i}$.$\overrightarrow{j}$ = $\overrightarrow{j}$.$\overrightarrow{i}$ = 0
or, -2a + 12 = 0
∴ a = 6

If $\overrightarrow{a}$ = 3$\overrightarrow{i}$ - m$\overrightarrow{j}$ and $\overrightarrow{b}$ = 10$\overrightarrow{i}$ + 6$\overrightarrow{j}$ are perpendicular to each other, what is the value of m?

Given,
$\overrightarrow{a}$ = 3$\overrightarrow{i}$ - m$\overrightarrow{j}$
$\overrightarrow{b}$ = 10$\overrightarrow{i}$ + 6$\overrightarrow{j}$

Since, $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$, $\overrightarrow{a} . \overrightarrow{b}$ = 0 or, (3$\overrightarrow{i}$ - m$\overrightarrow{j}$).(10$\overrightarrow{i}$ + 6$\overrightarrow{j}$) = 0
or, 30($\overrightarrow{i}$)2 + 18($\overrightarrow{i}.\overrightarrow{j}$) - 10m($\overrightarrow{j}.\overrightarrow{i}$) - 6m($\overrightarrow{j}$)2 = 0
We know, $\overrightarrow{i}$.$\overrightarrow{i}$ = 1, $\overrightarrow{j}$.$\overrightarrow{j}$ = 1, $\overrightarrow{i}$.$\overrightarrow{j}$ = $\overrightarrow{j}$.$\overrightarrow{i}$ = 0
or, 30 - 6m = 0
∴ m = 5