For 0, let be the angle between the vectors a=i+ j-3k and b=3i-j+2k.

Question

For $\lambda >0$, let $	heta $ be the angle between the vectors $\vec{a}=\hat{i}+\lambda \hat{j}-3\hat{k}$ and $\vec{b}=3\hat{i}-\hat{j}+2\hat{k}$. If the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are mutually perpendicular, then the value of (14 cos $	heta {)}^{2}$ is equal to

Accepted Answer

We are given the vectors

$\vec{a}=\hat{i}+\lambda \, \hat{j}-3\, \hat{k}\, \, \vec{b}=3\, \hat{i}-\hat{j}+2\, \hat{k},$

with $\lambda >0$, and we are told that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular. Our goal is to find $(14\cos 	heta {)}^{2}$, where $	heta $ is the angle between $\vec{a}$ and $\vec{b}$.

Use the Perpendicularity Condition: Two vectors are perpendicular if their dot product is zero. So, we require:

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

First, compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$:

$\vec{a}+\vec{b}=(\hat{i}+3\, \hat{i})+(\lambda \, \hat{j}-\hat{j})+(-3\, \hat{k}+2\, \hat{k})=4\, \hat{i}+(\lambda -1)\, \hat{j}-\hat{k}.$

$\vec{a}-\vec{b}=(\hat{i}-3\, \hat{i})+(\lambda \, \hat{j}+\hat{j})+(-3\, \hat{k}-2\, \hat{k})=-2\, \hat{i}+(\lambda +1)\, \hat{j}-5\, \hat{k}.$

Now, take their dot product:

$(4\, \hat{i}+(\lambda -1)\, \hat{j}-\hat{k})\cdot (-2\, \hat{i}+(\lambda +1)\, \hat{j}-5\, \hat{k})$

$=4(-2)+(\lambda -1)(\lambda +1)+(-1)(-5)=-8+({\lambda }^{2}-1)+5.$

$={\lambda }^{2}-4.$

Setting the dot product to zero:

${\lambda }^{2}-4=0\, \implies \, {\lambda }^{2}=4.$

Since $\lambda >0$, we have:

$\lambda =2.$

Find the Angle Between $\vec{a}$ and $\vec{b}$: Now that we know $\lambda =2$, the vectors become:

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

The cosine of the angle $	heta $ between $\vec{a}$ and $\vec{b}$ is given by:

$\cos 	heta =\frac{\vec{a}\cdot \vec{b}}{\parallel \vec{a}\parallel \parallel \vec{b}\parallel }.$

Calculate $\vec{a}\cdot \vec{b}$:

$\vec{a}\cdot \vec{b}=(1)(3)+(2)(-1)+(-3)(2)=3-2-6=-5.$

Calculate the magnitudes:

$\parallel \vec{a}\parallel =\sqrt{{1}^{2}+{2}^{2}+(-3{)}^{2}}=\sqrt{1+4+9}=\sqrt{14}.$

$\parallel \vec{b}\parallel =\sqrt{{3}^{2}+(-1{)}^{2}+{2}^{2}}=\sqrt{9+1+4}=\sqrt{14}.$

Thus,

$\cos 	heta =\frac{-5}{\sqrt{14}\cdot \sqrt{14}}=\frac{-5}{14}.$

Compute $(14\cos 	heta {)}^{2}$: $(14\cos 	heta {)}^{2}={(14\cdot \frac{-5}{14})}^{2}=(-5{)}^{2}=25.$

Final Answer
The value of $(14\cos 	heta {)}^{2}$ is:

$25.$

Solutions · 1