Vector Algebra

We are given two conditions:

The particles are at the same distance from the origin. This means that the magnitudes of their position vectors are equal.

The vectors are perpendicular (at right angles) to each other. This means their dot product is zero.

Let’s work through these step-by-step.

● The vectors are given by:

$\vec{A} = 2\hat{i} + 3n \hat{j} + 2\hat{k},$

$\vec{B} = 2\hat{i} -2\hat{j} + 4p \hat{k}.$

● Since the vectors are perpendicular, their dot product must be zero:

$\vec{A} \cdot \vec{B} = (2)(2) + (3n)(-2) + (2)(4p) = 4 - 6n + 8p = 0.$

This can be rearranged to:

$8p = 6n - 4 \quad \Longrightarrow \quad p = \frac{3n - 2}{4} \quad \text{(Equation 1)}.$

● Since the distances from the origin are equal, the magnitudes of the vectors must be equal. Compute the squares of the magnitudes:

For $\vec{A}: \quad |\vec{A}|^2 = 2^2 + (3n)^2 + 2^2 = 4 + 9n^2 + 4 = 8 + 9n^2.$

For $\vec{B}: \quad |\vec{B}|^2 = 2^2 + (-2)^2 + (4p)^2 = 4 + 4 + 16p^2 = 8 + 16p^2.$

Setting them equal:

$8 + 9n^2 = 8 + 16p^2 \quad \Longrightarrow \quad 9n^2 = 16p^2.$

This simplifies to:

$p^2 = \frac{9n^2}{16} \quad \Longrightarrow \quad p = \pm \frac{3n}{4} \quad \text{(Equation 2)}.$

● Now equate the two expressions for $p$ from Equation 1 and Equation 2.

Case 1: Assume $p = \frac{3n}{4}.$

Then,

$\frac{3n - 2}{4} = \frac{3n}{4} \quad \Longrightarrow \quad 3n - 2 = 3n \quad \Longrightarrow \quad -2 = 0,$

which is a contradiction.

Case 2: Assume $p = -\frac{3n}{4}.$

Then,

$\frac{3n - 2}{4} = -\frac{3n}{4} \quad \Longrightarrow \quad 3n - 2 = -3n.$

Simplify by adding $3n$ to both sides:

$6n - 2 = 0 \quad \Longrightarrow \quad 6n = 2 \quad \Longrightarrow \quad n = \frac{1}{3}.$

● Since we found $n = \frac{1}{3},$ its reciprocal is:

$n^{-1} = 3.$

Thus, the value of $n^{-1}$ is $3.$

To solve this problem, we'll analyze the conditions given about the vectors $\vec{A}$ and $\vec{B}$, and their resultant $\vec{R}$.

Given:

1. $\vec{R} = \vec{A} + \vec{B}$ is perpendicular to $\vec{A}$.


2. The magnitude of $\vec{R}$ is half that of $\vec{B}$: $|\vec{R}| = \frac{1}{2}|\vec{B}|$.

Approach:

We know that if $\vec{R}$ is perpendicular to $\vec{A}$, then their dot product is zero:

$ \vec{R} \cdot \vec{A} = 0 $

Substitute $\vec{R} = \vec{A} + \vec{B}$:

$ (\vec{A} + \vec{B}) \cdot \vec{A} = 0 $

$ \Rightarrow $$ \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} = 0 $

$ \Rightarrow $$ |\vec{A}|^2 + |\vec{A}||\vec{B}|\cos \theta = 0 $

$ \Rightarrow $$ \cos \theta = -\frac{|\vec{A}|^2}{|\vec{A}||\vec{B}|} $

$ \Rightarrow $$ \cos \theta = -\frac{|\vec{A}|}{|\vec{B}|} $

Next, we use the second condition involving magnitudes:

$ |\vec{R}| = |\vec{A} + \vec{B}| = \frac{1}{2}|\vec{B}| $

$ \Rightarrow $$ \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos \theta} = \frac{1}{2}|\vec{B}| $

$ \Rightarrow $$ |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\frac{|\vec{A}|}{|\vec{B}|} = \frac{1}{4}|\vec{B}|^2 $

$ \Rightarrow $$ |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}|^2 = \frac{1}{4}|\vec{B}|^2 $

$ \Rightarrow $$ |\vec{B}|^2 - |\vec{A}|^2 = \frac{1}{4}|\vec{B}|^2 $

$ \Rightarrow $$ \frac{3}{4}|\vec{B}|^2 = |\vec{A}|^2 $

$ \Rightarrow $$ |\vec{A}| = \frac{\sqrt{3}}{2}|\vec{B}| $

Finally, substitute this back into the cosine formula:

$ \cos \theta = -\frac{\frac{\sqrt{3}}{2}|\vec{B}|}{|\vec{B}|} $

$ \cos \theta = -\frac{\sqrt{3}}{2} $

This value of $\cos \theta$ corresponds to an angle of $150^\circ$ because $\cos 150^\circ = -\frac{\sqrt{3}}{2}$.

Answer:

The angle between $\vec{A}$ and $\vec{B}$ is $150^\circ$.

To solve this problem, we will use the concepts of vector addition and magnitudes involving the dot product. Given the angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}\right)$, we can use the properties of dot products and magnitudes to find the required integer value of $n$.

First, let's use the condition $|\vec{a} + \vec{b}| = \sqrt{2} |\vec{a} - \vec{b}|$. Square both sides to remove the square roots:

$ |\vec{a} + \vec{b}|^2 = 2 |\vec{a} - \vec{b}|^2 $

Now we will expand both sides using the formula for the magnitude of the sum and difference of vectors:

$ |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} $

And

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

Substitute these back into the original equation:

$ \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 2 (\vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}) $

Expand the right-hand side:

$ \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 2 \vec{a} \cdot \vec{a} - 4 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{b} $

Rearrange all terms to one side to combine like terms:

$ \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{a} + 4 \vec{a} \cdot \vec{b} - 2 \vec{b} \cdot \vec{b} = 0 $

Combine like terms:

$ - \vec{a} \cdot \vec{a} + 6 \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{b} = 0 $

We know that:

$ \vec{a} \cdot \vec{a} = |\vec{a}|^2 \quad \text{and} \quad \vec{b} \cdot \vec{b} = |\vec{b}|^2 \quad \text{and} \quad \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) $

Since the angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}\right)$, we have:

$ \cos(\theta) = \frac{5}{9} $

Substitute these back into the equation:

$ -|\vec{a}|^2 + 6 |\vec{a}| |\vec{b}| \left(\frac{5}{9}\right) - |\vec{b}|^2 = 0 $

Simplify it further:

$ -|\vec{a}|^2 + \frac{30}{9} |\vec{a}| |\vec{b}| - |\vec{b}|^2 = 0 $

$ -|\vec{a}|^2 + \frac{10}{3} |\vec{a}| |\vec{b}| - |\vec{b}|^2 = 0 $

We know that $|\vec{a}| = n |\vec{b}|$. Substitute this into the equation:

$ -(n |\vec{b}|)^2 + \frac{10}{3} (n |\vec{b}|) |\vec{b}| - |\vec{b}|^2 = 0 $

Simplify it:

$ -n^2 |\vec{b}|^2 + \frac{10}{3} n |\vec{b}|^2 - |\vec{b}|^2 = 0 $

Factor out $|\vec{b}|^2$:

$ |\vec{b}|^2 \left(-n^2 + \frac{10}{3} n - 1\right) = 0 $

Since $|\vec{b}|^2 \neq 0$, we can solve:

$ -n^2 + \frac{10}{3} n - 1 = 0 $

Multiply through by 3 to clear the fraction:

$ -3n^2 + 10n - 3 = 0 $

Rearrange it to match standard quadratic form:

$ 3n^2 - 10n + 3 = 0 $

Solve this quadratic equation using the quadratic formula:

$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

Here, $a = 3$, $b = -10$, and $c = 3$:

$ n = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} $

This results in two possible solutions for $n$:

$ n = \frac{18}{6} = 3 \quad \text{and} \quad n = \frac{2}{6} = \frac{1}{3} $

However, since $n$ is given to be an integer, we take:

$ \boxed{3} $

$\begin{aligned} &\begin{aligned} & (\vec{P}+\vec{Q})=\sqrt{2} A \\ & R=A \\ & \theta=90^{\circ} \end{aligned}\\ &\text { Resultant }=\sqrt{3} A \end{aligned}$

To determine the value of $ x $, given the vectors $\vec{A}=(-x \hat{i} - 6 \hat{j} - 2 \hat{k})$, $\vec{B}=(-\hat{i} + 4 \hat{j} + 3 \hat{k})$, and $\vec{C}=(-8 \hat{i} - \hat{j} + 3 \hat{k})$, and the condition $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$, we proceed as follows:

First, we calculate the cross product $\vec{B} \times \vec{C}$:

$ \vec{B} \times \vec{C} = 15 \hat{i} - 21 \hat{j} + 33 \hat{k} $

Next, using the condition $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$, we compute the dot product:

$ \vec{A} \cdot (\vec{B} \times \vec{C}) = (-x \hat{i} - 6 \hat{j} - 2 \hat{k}) \cdot (15 \hat{i} - 21 \hat{j} + 33 \hat{k}) $

$ \Rightarrow (-x)(15) + (-6)(-21) + (-2)(33) = 0 $

$ \Rightarrow -15x + 126 - 66 = 0 $

Solving this equation for $ x $:

$ -15x + 60 = 0 $

$ \Rightarrow x = 4 $

Thus, the value of $ x $ is $ 4 $.

To find a vector that has the same magnitude as vector $\vec{A}$ and is parallel to vector $\vec{B}$, we use the formula:

$\vec{N} = |\vec{A}| \hat{B}$

First, let's find the magnitude of $\vec{A}$, which is $|\vec{A}|$.

$|\vec{A}| = \sqrt{3^2 + 4^2}=5$

We're given that $\vec{B} = 4 \hat{i}+3 \hat{j}$, so to make $\vec{N}$ parallel to $\vec{B}$ and have it have the same magnitude as $\vec{A}$, they should be the same when $\vec{N}$'s magnitude is divided by 5, since $|\vec{A}|=5$.

So, $\vec{N} = \frac{5(4\hat{i}+3\hat{j})}{5} = 4\hat{i}+3\hat{j}$.

This means the $x$ component of the vector in the first quadrant is 4, and the $y$ component is given as 3. So, $x=4$.

$\overrightarrow A = 2\widehat i + 4\widehat j - 2\widehat k$

$\overrightarrow B = \widehat i + 2\widehat j + \alpha \widehat k$

$\overrightarrow A \,.\,\overrightarrow B = 0$, as $\overrightarrow A $ should be perpendicular to $\overrightarrow B $

$ \Rightarrow 2 + 8 - 2\alpha = 0$

$\alpha = 5$

To solve the problem, we need to determine the vectors normal to the planes containing given vectors and find the angle between those normal vectors.

First, let's find the vectors normal to the plane containing vectors $\overrightarrow{A}$ and $\overrightarrow{B}$. The normal vector can be calculated using the cross product:

$\overrightarrow{A} \times \overrightarrow{B} = (\widehat{i} + \widehat{j}) \times (\widehat{j} + \widehat{k})$.

Using the properties of the cross product:

$\begin{aligned} \overrightarrow{A} \times \overrightarrow{B} &= (\widehat{i} \times \widehat{j}) + (\widehat{i} \times \widehat{k}) + (\widehat{j} \times \widehat{j}) + (\widehat{j} \times \widehat{k}) \\ &= \widehat{k} - \widehat{j} + 0 + \widehat{i} \\ &= \widehat{i} - \widehat{j} + \widehat{k}. \end{aligned}$

So, the vector normal to the plane containing vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\overrightarrow{N}_{1} = \widehat{i} - \widehat{j} + \widehat{k}$.

Next, let's find the vector normal to the plane containing vectors $\overrightarrow{A}$ and $\overrightarrow{C}$:

$\overrightarrow{A} \times \overrightarrow{C} = (\widehat{i} + \widehat{j}) \times (-\widehat{i} + \widehat{j})$.

Using the properties of the cross product:

$\begin{aligned} \overrightarrow{A} \times \overrightarrow{C} &= (\widehat{i} \times -\widehat{i}) + (\widehat{i} \times \widehat{j}) + (\widehat{j} \times -\widehat{i}) + (\widehat{j} \times \widehat{j}) \\ &= 0 + \widehat{k} - \widehat{k} + 0 \\ &= \widehat{k} - \widehat{k} + 0 \\ &= 2 \widehat{k}. \end{aligned}$

So, the vector normal to the plane containing vectors $\overrightarrow{A}$ and $\overrightarrow{C}$ is $\overrightarrow{N}_{2} = 2 \widehat{k}$. But we just need the direction of this vector, not its magnitude, so we can simplify it to $\widehat{k}$.

Now, to find the angle between the two normal vectors $\overrightarrow{N}_{1} = \widehat{i} - \widehat{j} + \widehat{k}$ and $\overrightarrow{N}_{2} = \widehat{k}$, we use the dot product formula:

$\cos(\theta) = \frac{\overrightarrow{N}_{1} \cdot \overrightarrow{N}_{2}}{||\overrightarrow{N}_{1}|| ||\overrightarrow{N}_{2}||}$

First, calculate the dot product:

$\overrightarrow{N}_{1} \cdot \overrightarrow{N}_{2} = (\widehat{i} - \widehat{j} + \widehat{k}) \cdot \widehat{k} = 0 + 0 + 1 = 1$

Next, find the magnitude of the vectors:

$||\overrightarrow{N}_{1}|| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$

$||\overrightarrow{N}_{2}|| = \sqrt{0^2 + 0^2 + 1^2} = 1$

Now substitute these into the cosine formula:

$\cos(\theta) = \frac{1}{\sqrt{3} \times 1} = \frac{1}{\sqrt{3}} = \sqrt{\frac{1}{3}}$

Therefore, $\cos^{-1}(\sqrt{\frac{1}{3}}) = \cos^{-1}(\frac{1}{\sqrt{3}})$ indicates $x = 3$.

So, the value of $x$ is:

3