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

Hint: 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