Edexcel A-Level Mathematics A2 P4.7.2 Magnitude of a Vector QuestionsPractise using vector magnitude to find lengths, unit vectors, distances from points and triangle areas in 2D or 3D.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointscalculate |a| from components and use it for exact distances or simplified surdslink magnitudes to triangle areas, closest-point problems or parameter equationsfind a unit vector by dividing a direction vector by its magnitude
Edexcel A-Level Mathematics A2 P4.7.2 Magnitude of a Vector Questions question 1[Maximum number: 3]With respect to a fixed origin O, the line l1l_{1}l1 is given by the equationr=i+2j+5k+λ(8i−j+4k)\mathbf{r}=\mathbf{i}+2 \mathbf{j}+5 \mathbf{k}+\lambda(8 \mathbf{i}-\mathbf{j}+4 \mathbf{k})r=i+2j+5k+λ(8i−j+4k)where λ\lambdaλ is a scalar parameter.The point A lies on l1l_{1}l1Given that ∣OA→∣=510|\overrightarrow{O A}|=5 \sqrt{10}∣OA∣=510show that at A the parameter λ\lambdaλ satisfies81λ2+52λ−220=081 \lambda^{2}+52 \lambda-220=081λ2+52λ−220=0Mark as masteredShow Answer∣OA→∣2=(1+8λ)2+(2−λ)2+(5+4λ)2|\overrightarrow{OA}|^2=(1+8\lambda)^2+(2-\lambda)^2+(5+4\lambda)^2∣OA∣2=(1+8λ)2+(2−λ)2+(5+4λ)2∣OA→∣=510⇒(1+8λ)2+(2−λ)2+(5+4λ)2=250|\overrightarrow{OA}|=5\sqrt{10} \Rightarrow (1+8\lambda)^2+(2-\lambda)^2+(5+4\lambda)^2=250∣OA∣=510⇒(1+8λ)2+(2−λ)2+(5+4λ)2=25064λ2+16λ+1+λ2−4λ+4+16λ2+40λ+25=25081λ2+52λ−220=0\begin{aligned} 64\lambda^2+16\lambda+1+\lambda^2-4\lambda+4+16\lambda^2+40\lambda+25&=250\\ 81\lambda^2+52\lambda-220&=0 \end{aligned}64λ2+16λ+1+λ2−4λ+4+16λ2+40λ+2581λ2+52λ−220=250=0A1*Add to Test