Let =4 i+3 j+5 k and = i+2 j-4 k.

Question

Let $\vec\alpha=4\hat i+3\hat j+5\hat k$ and $\vec\beta=\hat i+2\hat j-4\hat k$. Let $\vec\beta_1$ be parallel to $\vec\alpha$ and $\vec\beta_2$ perpendicular to $\vec\alpha$, with $\vec\beta=\vec\beta_1+\vec\beta_2$. Then $5\vec\beta_2\cdot(\hat i+\hat j+\hat k)$ is:

Accepted Answer

$\vec\beta_1=\lambda\vec\alpha$, $\vec\beta_2=\vec\beta-\lambda\vec\alpha = (1-4\lambda)\hat i+(2-3\lambda)\hat j-(5\lambda+4)\hat k$.

$\vec\beta_2\cdot\vec\alpha=0$: $4(1-4\lambda)+3(2-3\lambda)-5(5\lambda+4)=0 \implies -50\lambda-10=0 \implies \lambda=-\dfrac15$.

$$\vec\beta_2 = \left(\frac95,\frac{13}5,-3\right) \implies 5\vec\beta_2 = (9,13,-15)$$
$$5\vec\beta_2\cdot(\hat i+\hat j+\hat k) = 9+13-15 = 7$$

Solutions · 1