First slide

Algebra of vectors

Question

Find the value of $λ$ so that the points P, Q, R and S on the sides OA, OB, OC and AB, respectively, of a regular tetrahedron OABC are coplanar. It is given $\frac{OP}{OA} = \frac{1}{3}, \frac{OQ}{OB} = \frac{1}{2}, \frac{OR}{OC} = \frac{1}{3} and \frac{OS}{AB} = λ$

Difficult

A

$λ = \frac{1}{2}$
B

$λ = - 1$
C

$λ = 0$
D

for no value of $λ$

Solution

$\begin{array}{l} Let \vec{OA} = \vec{a}, \vec{OB} = \vec{b} and \vec{OC} = \vec{c} \\ then \vec{AB} = \vec{b} - \vec{a} and \vec{OP} = \frac{1}{3} \vec{a} \\ \vec{OQ} = \frac{1}{2} \vec{b}, \vec{OR} = \frac{1}{3} \vec{c} \end{array}$
Since P, Q,, R and S are coplanar, then $\vec{PS} = α \vec{PQ} + β \vec{PR}$
( $\vec{PS}$ can be written as a linear combination of $\vec{PQ}$ and $\vec{PR}$ )
$\begin{array}{l} = α (\vec{OQ} - \vec{OP}) + β (\vec{OR} - \vec{OP}) \\ \vec{OS} - \vec{OP} = - (α + β) \frac{\vec{a}}{3} + \frac{α}{2} \vec{b} + \frac{β}{3} \vec{c} \\ \Rightarrow \vec{OS} = (1 - α - β) \frac{\vec{a}}{3} + \frac{α}{2} \vec{b} + \frac{β}{3} \vec{c} - - - (i) \\ given \vec{OS} = λ \vec{AB} = λ (\vec{b} - \vec{a}) - - - (ii) \end{array}$
$From (i) and (ii), β = 0, \frac{1 - α}{3} = - λ and \frac{α}{2} = λ$
$\begin{aligned} \Rightarrow 2 λ = 1 + 3 λ \\ or λ = - 1 \end{aligned}$

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