OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio 1:3. AC and MQ meet at P. If O to A = a, O to C = c . Express O to P in terms of a and c.
1. If →Q=→c+34→a and →M=12→a so line ¯MQ is 12→a+λ(→Q−→M)=12→a+λ(→c+14→a)
2. Line ¯AC is →a+μ(→c−→a)
3.
The intersection of the line ¯MQ with the line ¯AC is $$ the point →P=12→a+λ(→c+14→a).$$ We equate: 12→a+λ(→c+14→a)=→a+μ(→c−→a) $$ so λ(→c+14→a)−μ(→c−→a)=12→a $$ The perpendicular vector of (→c−→a) is (→c+→a), because we have a square and $$ (→c−→a)∗(→c+→a)=0, because the dot-product of vectors $$ perpendicular to each other is a Zero. $$ We multiply our equation with (→c+→a) and see what passed:
$\\$\small{\text{ $ \lambda (\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a}) - \mu \underbrace{( \vec{c}-\vec{a}) (\vec{c}+\vec{a})}_{dot-product = 0} = \frac{1}{2}\vec{a} (\vec{c}+\vec{a}) $ }} $\\$\small{\text{ Now we can dissolve after $\lambda$ and receive for $\lambda$: }} $\\$\small{\text{ $ \lambda = \dfrac{ \frac{1}{2}\vec{a} (\vec{c}+\vec{a}) } {(\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a}) } $ }} $\\$ \boxed{ \small{\text{ $\vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $}} = \lambda \vec{c} + (\frac{1}{2}+\frac{\lambda}{4})\vec{a} $ }}}
