Question

Obtain the projection of a vector Q=4i+6j-k along a line L which originate at point
(2,2,0) and passes through point (-3,4,5).

Answer 1

Here,

$\longrightarrow\vec{\sf{a}}=\sf{4\ \hat i+6\ \hat j-\hat k}$

The vector parallel to the line passing through the points (2, 2, 0) and (-3, 4, 5) can be obtained as,

$\longrightarrow\vec{\sf{b}}=\sf{(-3-2)\ \hat i+(4-2)\ \hat j+(5-0)\ \hat k}$

$\longrightarrow\vec{\sf{b}}=\sf{-5\ \hat i+2\ \hat j+5\ \hat k}$

Let's find magnitude of each vector.

$\sf{\longrightarrow a=\sqrt{4^2+6^2+(-1)^2}}$

$\sf{\longrightarrow a=\sqrt{53}}$

and,

$\sf{\longrightarrow b=\sqrt{(-5)^2+2^2+5^2}}$

$\sf{\longrightarrow b=\sqrt{54}}$

Let $\theta$ be the angle between $\vec{\sf{a}}$ and $\vec{\sf{b}}$ whose cosine is given by,

$\longrightarrow\mathsf{\cos\theta}=\dfrac{\vec{\sf{a}}\cdot\vec{\sf{b}}}{\mathsf{ab}}$

$\sf{\longrightarrow\cos\theta=\dfrac{(4\ \hat i+6\ \hat j-\hat k)\cdot(-5\ \hat i+2\ \hat j+5\ \hat k)}{\sqrt{53\times54}}}$

$\sf{\longrightarrow\cos\theta=\dfrac{-20+12-5}{\sqrt{53\times54}}}$

$\sf{\longrightarrow\cos\theta=-\dfrac{13}{\sqrt{53\times54}}}$

Hence, projection of $\vec{\sf{a}}$ over the line is,

$\longrightarrow\vec{\sf{a_b}}=\sf{a\cos\theta\ \hat b}$

$\longrightarrow\vec{\sf{a_b}}=\sf{\sqrt{53}\times-\dfrac{13}{\sqrt{53\times54}}\left(\dfrac{-5\ \hat i+2\ \hat j+5\ \hat k}{\sqrt{54}}\right)}$

$\longrightarrow\underline{\underline{\vec{\sf{a_b}}=\sf{-\dfrac{13}{54}\left(-5\ \hat i+2\ \hat j+5\ \hat k\right)}}}$