Question

ans fast and get 5.0 star...​

Answer 1

Answer:

49

Step-by-step explanation:

I hope it will help you understand the concept and bye

Answer 2

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

$\sf \: Let \: \vec{a} \: = \: 2\hat i \: + 2\hat j \: - 3\hat k \:$

$\sf \: Let \: \vec{b} \: = \: 3\hat i \: + k\hat j \: + 2\hat k \:$

$\sf \: Let \: \vec{c} \: = \: \hat i + 2\hat j \: + 3\hat k \:$

It is given that

• Vectors are co-planar,

$\sf \: \therefore \: \bf \: [ \: \vec{a} \: \: \vec{b} \: \: \vec{c} \: ] = \: 0$

$\sf \: \begin{array}{|ccc|}\sf 2 &\sf 2 &\sf - 3 \\ \sf 3 &\sf k &\sf 2 \\ \sf 1 &\sf 2 &\sf 3\end{array} = 0$

$\sf \: 2(3k - 4) - 2(9 - 2) - 3(6 - k) = 0$

$\sf \: 6k - 8 - 14 - 18 + 3k = 0$

$\sf \: 9k - 40 = 0$

$\: \: \: \boxed{ \bf{k \: = \: \dfrac{40}{9} }}$

$\: \boxed{ \bf{Hence, \: option \: (B) \: is \: correct}}$

Additional Information :-

$1. \: \: \: \boxed{ \bf{[ \: \vec{a} \: \: \vec{b} \: \: \vec{c} \: ] = \vec{a}.(\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}).\vec{c}}}$

$2. \: \: \: \boxed{ \bf{[ \: \vec{a} \: \: \vec{b} \: \: \vec{c} \: ] = [ \: \vec{b} \: \: \vec{c} \: \: \vec{a} \: ] = [ \: \vec{c} \: \: \vec{a} \: \: \vec{b} \: ]}}$

$3. \: \: \: \boxed{ \bf{[ \: \vec{a} \: \: \vec{b} \: \: \vec{b} \: ] = 0}}$

$4. \: \: \: \boxed{ \bf{[ \: \vec{a} + \vec{d} \: \: \vec{b} \: \: \vec{c} \: ] = [ \: \vec{a} \: \: \vec{b} \: \: \vec{c} \: ] + [ \: \vec{d} \: \: \vec{b} \: \: \vec{c} \: ]}}$