Question

solve this ques of jee mains
chapter → vectors​

Answer 1

Answer:

(b) -1 is the answer.

refer to attachment

Answer 2

$\large\underline{\sf{Solution-}}$

Given that the following vectors,

$\rm :\longmapsto\:(1,a, {a}^{2}), \: (1,b, {b}^{2}), \: (1,c, {c}^{2}) \: are \: non \: coplanar$

So, it implies,

$\rm :\longmapsto\:\rm \: \: \: \: \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered} \: \ne \: 0 - - - (1)$

Now, given that

$\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}a& {a}^{2} &1 + {a}^{3} \\b& {b}^{2} &1 + {b}^{3} \\c& {c}^{2} &1 + {c}^{3} \end{array}\right | \end{gathered} = 0$

Using splitting property of determinants, if any row or column is represented as a sum of two or more elements, then determinant can be expressed as a sum of two or more determinants.

So, we have now

$\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}a& {a}^{2} &{a}^{3} \\b& {b}^{2} &{b}^{3} \\c& {c}^{2} &{c}^{3} \end{array}\right | \end{gathered} +\begin{gathered}\sf \left | \begin{array}{ccc}a& {a}^{2} &1 \\b& {b}^{2} &1 \\c& {c}^{2} &1 \end{array}\right | \end{gathered} = 0$

Now, take out a, b, c common from respective Rows, we get

$\rm :\longmapsto\:abc\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered} +\begin{gathered}\sf \left | \begin{array}{ccc}a& {a}^{2} &1 \\b& {b}^{2} &1 \\c& {c}^{2} &1 \end{array}\right | \end{gathered} = 0$

Now, we know, if successive rows or columns are interchanged, the determinant value is multiplied by - 1.

So, in second determinant, interchanged second column with third column, so we have

$\rm :\longmapsto\:abc\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered} - \begin{gathered}\sf \left | \begin{array}{ccc}a&1 & {a}^{2} \\b&1 & {b}^{2} \\c&1 & {c}^{2} \end{array}\right | \end{gathered} = 0$

Again, interchanged first column with second column, we get

$\rm :\longmapsto\:abc\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered} + \begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered} = 0$

can be rewritten as after taking common,

$\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered}(1 + abc) = 0$

Since, from equation (1) we have

$\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}1&a& {a}^{2} \\1&b& {b}^{2} \\1&c& {c}^{2} \end{array}\right | \end{gathered} \: \ne \: 0$

$\rm :\implies\:1 + abc = 0$

$\bf\implies \:abc \: = \: - \: 1$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: Option \: (b) \: is \: correct}}}$

Additional Information :-

$\underbrace{\boxed{ \tt{ [\vec{a} \: \: \vec{b} \: \: \vec{c}]\: = \: \vec{a}. \: (\vec{b} \times \vec{c})}}}$

$\underbrace{\boxed{ \tt{ [\vec{a} \: \: \vec{a} \: \: \vec{b}]\: = 0}}}$