Given points A (2, 1, -1), B(3, 2, -1), and C (3, 1, 0). Find the angle between the vectors $\to A B a n d \to A C$ .

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Solution

$A (2, 1, - 1), B (3, 2, - 1), C (3, 1, 0)$
$¯ ¯¯¯¯¯¯ ¯ A B = (3 - 2)^i + (2 - 1)^j + (- 1 - (- 1))^k$
$=^i +^j$
$¯ ¯¯¯¯¯¯ ¯ A C = (3 - 3)^i + (1 - 2)^j + (0 - (- 1))^k$
$=^- j +^k$
Angle bet $^{n}$ $¯ ¯¯¯¯¯¯ ¯ A B$ & $¯ ¯¯¯¯¯¯ ¯ A C$ ,
$θ = c o s^{- 1} ⎛ ⎝ ∣ ∣ ∣ ∣ \frac{¯ ¯¯¯¯¯¯ ¯ A B, ¯ ¯¯¯¯¯¯ ¯ A C}{| ¯ ¯¯¯¯¯¯ ¯ A B | | ¯ ¯¯¯¯¯¯ ¯ A C |} ∣ ∣ ∣ ∣ ⎞ ⎠$
$= c o s^{- 1} ⎛ ⎝ ∣ ∣ ∣ ∣ \frac{(^i +^j) (-^j +^k)}{(\sqrt{1^{2} + 1^{2}} \sqrt{1^{2} + 1^{2}})} ∣ ∣ ∣ ∣ ⎞ ⎠$
$= c o s^{1} (∣ ∣ ∣ \frac{- 1}{\sqrt{2} \sqrt{2}} ∣ ∣ ∣) = c o s^{- 1} (\frac{1}{2})$
$∴ θ = \frac{π}{3}$ OR $60^{\circ}$

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Given points A (2, 1, -1), B(3, 2, -1), and C (3, 1, 0). Find the angle between the vectors →ABand→AC.

Given points A (2, 1, -1), B(3, 2, -1), and C (3, 1, 0). Find the angle between the vectors $\to A B a n d \to A C$ .