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Question: If \( \overline a = 5\widehat i - \widehat j + 7\widehat k \) and \( \overline b = \widehat i - \wid...

If a‾=5i^−j^+7k^\overline a = 5\widehat i - \widehat j + 7\widehat k and b‾=i^−j^+λk^\overline b = \widehat i - \widehat j + \lambda \widehat k , find λ\lambda for which a‾+b‾\overline a + \overline b and a‾−b‾\overline a - \overline b

Explanation

Solution

Hint : Here we are given the vectors which are perpendicular means the product is zero, so will find the product and then will simplify for the required value of remember the term perpendicular or the orthogonal means the product is equal to zero.

Complete step-by-step answer :
Given that: a‾+b‾\overline a + \overline b and a‾−b‾\overline a - \overline b are perpendicular, then its dot product will be zero.
⇒(a‾+b‾).(a‾−b‾)=0\Rightarrow (\overline a + \overline b ).(\overline a - \overline b ) = 0
Expand the brackets finding the product of the terms.
⇒(a‾).(a‾)−(a‾).(b‾)+(a‾).(b‾)−(b‾).(b‾)=0\Rightarrow (\overline a ).(\overline a ) - (\overline a ).(\overline b ) + (\overline a ).(\overline b ) - (\overline b ).(\overline b ) = 0
Like terms with the same values and opposite signs cancels each other.
⇒(a‾)2−(b‾)2=0\Rightarrow {(\overline a )^2} - {(\overline b )^2} = 0
Move one term on the opposite side, when you move any term from one side to another the sign to the opposite side then the sign of the terms also changes. Positive term changes to negative and vice-versa.
⇒(a‾)2=(b‾)2\Rightarrow {(\overline a )^2} = {(\overline b )^2}
Take square roots on both sides of the equation.
⇒(a‾)2=(b‾)2\Rightarrow \sqrt {{{(\overline a )}^2}} = \sqrt {{{(\overline b )}^2}}
Square and square-root cancel each other out.
⇒∣a‾∣=∣b‾∣\Rightarrow \left| {\overline a } \right| = \left| {\overline b } \right|
Place the values from the given data where a‾=5i^−j^+7k^\overline a = 5\widehat i - \widehat j + 7\widehat k and b‾=i^−j^+λk^\overline b = \widehat i - \widehat j + \lambda \widehat k
⇒(5)2+(−1)2+(7)2=(1)2+(−1)2+(λ)2\Rightarrow \sqrt {{{(5)}^2} + {{( - 1)}^2} + {{(7)}^2}} = \sqrt {{{(1)}^2} + {{( - 1)}^2} + {{(\lambda )}^2}}
Simplify finding the sum of the squares. Square of any positive or negative term is always positive.
⇒25+1+49=1+1+λ2\Rightarrow \sqrt {25 + 1 + 49} = \sqrt {1 + 1 + {\lambda ^2}}
Simplify the above expression finding the sum of the terms inside the square-root.
⇒75=2+λ2\Rightarrow \sqrt {75} = \sqrt {2 + {\lambda ^2}}
Take square on both the sides of the equation,
⇒(75)2=(2+λ2)2\Rightarrow {\left( {\sqrt {75} } \right)^2} = {\left( {\sqrt {2 + {\lambda ^2}} } \right)^2}
Square and square root cancels each other on both the sides of the equation –
⇒75=2+λ2\Rightarrow 75 = 2 + {\lambda ^2}
Make lambda the subject and move constants on the opposite side. When you move any term from one side of the equation to the opposite side then the sign of the term changes. Positive term becomes negative and vice-versa.
⇒75−2=λ2\Rightarrow 75 - 2 = {\lambda ^2}
Simplify –
⇒λ2=73\Rightarrow {\lambda ^2} = 73
Take square-root on both the sides of the equation –
⇒λ=±73\Rightarrow \lambda = \pm \sqrt {73}
This is the required solution.
So, the correct answer is “ λ=±73\lambda = \pm \sqrt {73} ”.

Note : Always remember that square of any positive or negative term gives positive term whereas square root of any positive term gives positive or negative term that’s why we have kept plus or minus sign at the end of the solution taking the square-root of the term.