Let a=4 i+3 j and b=3 i-4 j+5 k, and c be a vector such that c( a b)+25=0, c( i+ j+ k)=4, …

Question

Let $\vec a=4\hat i+3\hat j$ and $\vec b=3\hat i-4\hat j+5\hat k$, and $\vec c$ be a vector such that $\vec c\cdot(\vec a	imes\vec b)+25=0$, $\vec c\cdot(\hat i+\hat j+\hat k)=4$, and the projection of $\vec c$ on $\vec a$ is $1$. Then the projection of $\vec c$ on $\vec b$ equals:

Accepted Answer

$\vec a\times\vec b = (15,-20,-25)$. Let $\vec c=(x,y,z)$:

$\vec c\cdot(\vec a\times\vec b)=-25 \Rightarrow 15x-20y-25z=-25 \Rightarrow 3x-4y-5z=-5$

$x+y+z=4$

Projection on $\vec a$ is 1: $\dfrac{4x+3y}{5}=1 \Rightarrow 4x+3y=5$

Solving these three equations: $x=2,\ y=-1,\ z=3$, so $\vec c = 2\hat i-\hat j+3\hat k$.

$$\text{Projection on }\vec b = \frac{\vec c\cdot\vec b}{|\vec b|} = \frac{6+4+15}{\sqrt{50}} = \frac{25}{5\sqrt2} = \frac5{\sqrt2}$$

Solutions · 1