 # Doubt from vector

#1
#2

according to given options answer should be 4i.e option c..
Hope it is all clear now

#3 I am unable to understand the reason from where the condition like z+x=0 all that came

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#4

Look left hand side (z+x)(a×b) is present but loo k right hand side
No term contain a×b so (z+x)=0

#5

@darshil_2020

Whenever you are dealing with vectorial equations, just as complex numbers you can compare the left and right hand sides.

Quoting from your posted solution :

(2+x)\vec{a}+(2+y)\vec{b}+(z+x)(\vec{a}X\vec{b})=z(b^2\vec{a}-(\vec{a}.\vec{b})\vec{b})

So here, we can compare both the sides, the coefficients of \vec{a},\vec{b} and their cross product as they resemble independent entities.

Thus we will get : x+2=b^2z\space,\space y+2=-absin\theta z \space \& \space x+z=0 (as there is no coefficient of their cross product on the right hand side and here we have considered \theta to be the angle between \vec{a} and \vec{b})

Do let me know if you have any doubts.

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#6

Note that \vec{a}.\vec{b}=1

1. Dot with \displaystyle \vec{a} to get 2\vec{a}.\vec{b}+2\vec{b}.\vec{b}+\vec{b}.\vec{c}=0

so that \displaystyle \vec{b}.\vec{c}=-4

1. Dot with \vec{a} to get 10+ \vec{a}.\vec{c}=[\vec{b} \ \vec{c} \ \vec{a}]

2. Dot with \vec{b}\times \vec{c}

\displaystyle [\vec{c} \ \vec{a} \ \vec{b}]= [\vec{b} \ \vec{c} \ \vec{a}]= (\vec{b}\times \vec{c}).(\vec{a}\times \vec{b}) =\begin{vmatrix} \vec{a}.\vec{b} & \vec{b}.\vec{b} \\\vec{c}.\vec{a}& \vec{b}.\vec{c}\end{vmatrix}= \begin{vmatrix} 1 & 1 \\\vec{c}.\vec{a}& -4\end{vmatrix}

so that [\vec{b} \ \vec{c} \ \vec{a}]+\vec{c}.\vec{a}=-4

Hence [\vec{b} \ \vec{c} \ \vec{a}]=3

and hence \left| [\vec{a} \ \vec{c} \ \vec{b}] \right|=3

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