First slide

Algebra of vectors

Question

If $\vec{α} + \vec{β} + \vec{γ} = k \vec{δ} and \vec{β} + \vec{γ} + \vec{δ} = m \vec{α}, \vec{α} and \vec{δ}$ are non-collinear, then $\vec{α} + \vec{β} + \vec{γ} + \vec{δ}$ equal

Moderate

A

$k \vec{α}$
B

$m \vec{δ}$
C

$\vec{0}$
D

$(k + m) \vec{γ}$

Solution

$Given \vec{α} + \vec{β} + \vec{γ} = k \vec{δ} - - - - (i)$
$\vec{β} + \vec{γ} + \vec{δ} = m \vec{α} - - - - (ii)$
From(i), $\vec{α} + \vec{β} + \vec{γ} + \vec{δ} = (k + 1) \vec{δ} - - - (iii)$
From (ii), $\vec{α} + \vec{β} + \vec{γ} + \vec{δ} = (m + 1) \vec{α} - - - - (iv)$
From (iii) and (iv), we get
$(k + 1) \vec{δ} = (m + 1) \vec{α} - - - - - (v)$
Since $\vec{α}$ is not parallel to $\vec{δ}$ ,
From (v),k+1=0and m+ 1=0
$From (iii), \vec{α} + \vec{β} + \vec{γ} + \vec{δ} = \vec{0}$

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