Let (alpha, beta, gamma) be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2) on the line -> r=(-hat i+3 hat j+hat k)+lambda(2 hat i+3 hat j-hat k) Then the length of the projection of the vector alpha hat i+beta hat j+gamma hat k on the vector 6 hat i+2 hat j+3 hat k is

/ line r=(-hat i+3 hat j+hat k)+lambda(2 hat i+3 hat j-hat k) frac x+12=frac y-33=frac z-1-1=lambda Any general point P on the line is (2 lambda-1,3 lambda+3,-lambda+1) Let the given point is A(5,4,2) / line AP(2 lambda-6) hat i+(3 lambda-1) hat j+(-lambda-1) hat k Because / line AP perp^ r Line ( L) Therefore / line AP * (2 hat i+3 hat j-hat k)=0 2(2 lambda-6)+3(3 lambda-1)-1(-lambda-1)=0 => lambda=1 Therefore alpha=1 beta=6 gamma=0 Let the vector / line u=alpha hat i+beta hat j+gamma hat k / line u=hat i+6 hat j+ O hat k \ / line w =6 hat i+2 hat j+3 hat k So projection =frac| / line u * / line w|| / line w|=(18 / 7)

Let alpha beta gamma be the co ordinates of the foot of... | null (Mathematics)

\overline{\mathrm{r}}=(-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})\\

\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-1}{-1}=\lambda\\

Any general point P on the line is

\\(2 \lambda-1,3 \lambda+3,-\lambda+1)\\

Let the given point is

\mathrm{A}(5,4,2)\\

\overline{\mathrm{AP}}(2 \lambda-6) \hat{\mathrm{i}}+(3 \lambda-1) \hat{\mathrm{j}}+(-\lambda-1) \hat{\mathrm{k}}\\

\because \overline{\mathrm{AP}} \perp^{\mathrm{r}}

Line

(\mathrm{L})\\

\therefore \overline{\mathrm{AP}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})=0\\

2(2 \lambda-6)+3(3 \lambda-1)-1(-\lambda-1)=0\\

\Rightarrow \lambda=1\\

\therefore \alpha=1\\

\beta=6\\

\gamma=0\\

Let the vector

\overline{\mathrm{u}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}

\overline{\mathrm{u}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+\mathrm{O} \hat{\mathrm{k}}

\& \overline{\mathrm{w}} \\=6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\\

So projection

=\frac{|\overline{\mathrm{u}} \cdot \overline{\mathrm{w}}|}{|\overline{\mathrm{w}}|}=\frac{18}{7}

Explanation

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