Question 16: Motion in 2D & Vectors

Question

If $$\vec{a}$$ and $$\vec{b}$$ makes an angle $$\cos ^{-1}\left(\frac{5}{9}\right)$$ with each other, then $$|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|$$ for $$|\vec{a}|=n|\vec{b}|$$ The integer value of $$\mathrm{n}$$ is _________.

Accepted Answer

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 $$\cos^{-1}\left(\frac{5}{9}ight)$$, 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:
$$ \vec{a} \cdot \vec{a} = |\vec{a}|^2 \quad 	ext{and} \quad \vec{b} \cdot \vec{b} = |\vec{b}|^2 \quad 	ext{and} \quad \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(	heta) $$
Since the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\cos^{-1}\left(\frac{5}{9}ight)$$, we have:
$$ \cos(	heta) = \frac{5}{9} $$
Substitute these back into the equation:
$$ -|\vec{a}|^2 + 6 |\vec{a}| |\vec{b}| \left(\frac{5}{9}ight) - |\vec{b}|^2 = 0 $$
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 $$|\vec{b}|^2$$:
$$ |\vec{b}|^2 \left(-n^2 + \frac{10}{3} n - 1ight) = 0 $$
Since $$|\vec{b}|^2 
eq 0$$, 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 $$n$$:
$$ n = \frac{18}{6} = 3 \quad 	ext{and} \quad n = \frac{2}{6} = \frac{1}{3} $$
However, since $$n$$ is given to be an integer, we take:
$$ \boxed{3} $$

Motion in 2D & Vectors

If $\vec{a} and \vec{b} makes an angle \cos ^{-1}\left(\frac{5}{9}\right) with each other, then |\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}| for |\vec{a}|=n|\vec{b}| The integer value of \mathrm{n}$ is _________.

Choose the correct answer: