Question

If we have vetor $\overrightarrow{a}=i-j+7k$ and vector $\overrightarrow{b}=5i-j+\lambda k$, then find the value of $\lambda$, so that $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ are perpendicular vectors.

Answer 1

We are given two vectors $\overrightarrow{a}=i-j+7k$ and $\overrightarrow{b}=5i-j+\lambda k$ first we will calculate the value of $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$, which will be $\overrightarrow{a}+\overrightarrow{b}=6i-2j+(7+\lambda )k$ and $\overrightarrow{a}-\overrightarrow{b}=-4i+(7-\lambda )k$
Now it’s given that $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ are perpendicular vectors.it means we can write their dot product will be 0, so $(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=(6i-2j+(7+\lambda )k).(-4i+(7-\lambda )k)=0$ and on solving this equation we will get the value of $\overrightarrow{a}-\overrightarrow{b}=-4i+(7-\lambda )k$

Complete step-by-step solution:
Given two vectors $\overrightarrow{a}=i-j+7k$ and $\overrightarrow{b}=5i-j+\lambda k$ we are asked to find value of $\lambda$, given condition is that $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ are perpendicular vectors, which means dot product of $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ is 0. So, for that we first have to calculate value of $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$
While calculating $\overrightarrow{a}+\overrightarrow{b}$ we will add $i$ components together, $j$components together and similarly $k$ components together which results into $\overrightarrow{a}+\overrightarrow{b}=(i-j+7k)+(5i-j+\lambda k)=(1+5)i-(1+1)j+(7+\lambda )k)$
Which equals to $\overrightarrow{a}+\overrightarrow{b}=6i-2j+(7+\lambda )k$ similarly we apply same procedure for calculating value of $\overrightarrow{a}-\overrightarrow{b}$ , which results into.
$\overrightarrow{a}-\overrightarrow{b}=(i-j+7k)-(5i-j+\lambda k)=(1-5)i+(1-1)j+(7-\lambda )k$
Which equals to $\overrightarrow{a}-\overrightarrow{b}=-4i+(7-\lambda )k$
As given dot product of $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ is 0, applying this condition we can write as
$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=(6i-2j+(7+\lambda )k).(-4i+(7-\lambda )k)=0$
Considering unit vector property in mind that is $i.i={{i}^{2}}=1,j.j={{j}^{2}}=1,k.k={{k}^{2}}=1$
And $i.j=0,i.k=0,j.k=0$ (because unit vectors components are perpendicular to each other that’s why their dot product is 0)
Solving it and considering this property $i.j=0, i.k=0, j.k=0$ we get our expression as $6\times (-4){{i}^{2}}+(-2)\times 0{{j}^{2}}+(7+\lambda )(7-\lambda ){{k}^{2}}=0$
Now putting this property $i.i={{i}^{2}}=1,j.j={{j}^{2}}=1,k.k={{k}^{2}}=1$
Our expression will look like
$6\times (-4)+(-2)\times 0+(7+\lambda )(7-\lambda )=0$
Which on solving looks like $6\times (-4)+(7+\lambda )(7-\lambda )=0$
(applying formula $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$)
On further solving we get $-24+0+49-{{\lambda }^{2}}=0$
Now expression will look like ${{\lambda }^{2}}=25$
Hence value of $\lambda$ is 5 and -5.

Note: We can also solve it directly by taking dot product of $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ at initially
$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})$ which results to $(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})={{a}^{2}}-a.b+b.a-{{b}^{2}}$
Now we know that $a.b=b.a$ it means $(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})={{a}^{2}}-{{b}^{2}}$
Given $\overrightarrow{a}=i-j+7k$ and $\overrightarrow{b}=5i-j+\lambda k$
So ${{a}^{2}}={{1}^{2}}+{{1}^{2}}+{{7}^{2}}=1+1+49=51$ and similarly ${{b}^{2}}={{5}^{2}}+{{1}^{2}}+{{\lambda }^{2}}=25+1+{{\lambda }^{2}}=26+{{\lambda }^{2}}$
$(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})={{a}^{2}}-{{b}^{2}}=0$ and on solving we again get the same result $\lambda$ is 5 and -5