Triangle ABC has vertices with position vectors i+2j+2k, 2i+5j−2k and 5i+4j+k. Find (a) AC, (b) the cartesian…

Question

Triangle ABC has vertices with position vectors i+2j+2k, 2i+5j−2k and 5i+4j+k. Find (a) AC, (b) the cartesian equation of line AC, (c) (i) the perpendicular distance of point B from line AC, (ii) the area of triangle ABC.

ScanSolve AI Tutor · Accepted Answer

Given the position vectors of the vertices of triangle ABC:
$\vec{A} = \vec{i} + 2\vec{j} + 2\vec{k}$
$\vec{B} = 2\vec{i} + 5\vec{j} - 2\vec{k}$
$\vec{C} = 5\vec{i} + 4\vec{j} + \vec{k}$

a) Find $\vec{AC}$ .

Calculate the vector $\vec{AC}$ by subtracting the position vector of A from the position vector of C.
$\vec{AC} = \vec{C} - \vec{A}$ $\vec{AC} = (5\vec{i} + 4\vec{j} + \vec{k}) - (\vec{i} + 2\vec{j} + 2\vec{k})$ $\vec{AC} = (5-1)\vec{i} + (4-2)\vec{j} + (1-2)\vec{k}$ $\vec{AC} = \boxed{\mathbf{4\vec{i} + 2\vec{j} - \vec{k}}}$
b) Find the cartesian equation of line AC.
Use point A $(1, 2, 2)$ and the direction vector $\vec{AC} = 4\vec{i} + 2\vec{j} - \vec{k}$ to form the cartesian equation of the line.
The parametric equations are $x = 1+4t$ , $y = 2+2t$ , $z = 2-t$ .
Equating $t$ from each equation gives the cartesian form:
$\frac{x-1}{4} = \frac{y-2}{2} = \frac{z-2}{-1}$ $\boxed{\mathbf{\frac{x-1}{4} = \frac{y-2}{2} = \frac{z-2}{-1}}}$
c) (i) Find the perpendicular distance of point B from line AC.
Calculate the vector $\vec{AB}$ .
$\vec{AB} = \vec{B} - \vec{A}$ $\vec{AB} = (2\vec{i} + 5\vec{j} - 2\vec{k}) - (\vec{i} + 2\vec{j} + 2\vec{k})$ $\vec{AB} = (2-1)\vec{i} + (5-2)\vec{j} + (-2-2)\vec{k}$ $\vec{AB} = \vec{i} + 3\vec{j} - 4\vec{k}$
Calculate the cross product $\vec{AB} \times \vec{AC}$ .
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 3 & -4 \\ 4 & 2 & -1 \end{vmatrix}$ $= \vec{i}((3)(-1) - (-4)(2)) - \vec{j}((1)(-1) - (-4)(4)) + \vec{k}((1)(2) - (3)(4))$ $= \vec{i}(-3 + 8) - \vec{j}(-1 + 16) + \vec{k}(2 - 12)$ $= 5\vec{i} - 15\vec{j} - 10\vec{k}$
Calculate the magnitude of $\vec{AC}$ and $\vec{AB} \times \vec{AC}$ .
$|\vec{AC}| = \sqrt{4^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$ $|\vec{AB} \times \vec{AC}| = \sqrt{5^2 + (-15)^2 + (-10)^2} = \sqrt{25 + 225 + 100} = \sqrt{350} = 5\sqrt{14}$
Calculate the perpendicular distance $d$ from point B to line AC using the formula $d = \frac{|\vec{AB} \times \vec{AC}|}{|\vec{AC}|}$ .
$d = \frac{5\sqrt{14}}{\sqrt{21}} = \frac{5\sqrt{2 \times 7}}{\sqrt{3 \times 7}} = \frac{5\sqrt{2}}{\sqrt{3}}$
Rationalize the denominator:
$d = \frac{5\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{6}}{3}$ $\text{Perpendicular distance} = \boxed{\mathbf{\frac{5\sqrt{6}}{3}}}$
c) (ii) Find the area of triangle ABC.
Calculate the area of triangle ABC using the formula Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}|$ .
$\text{Area} = \frac{1}{2} (5\sqrt{14})$ $\text{Area} = \boxed{\mathbf{\frac{5\sqrt{14}}{2}}}$

Triangle ABC has vertices with position vectors i+2j+2k, 2i+5j-2k and 5i+4j+k. Find (a) AC, (b) the cartesian equation of line AC, (c) (i) the perpendicular distance of point B from line AC, (ii) the area of triangle ABC.

Step-by-step solution

Calculate the vector $\vec{AC}$ by subtracting the position vector of A from the position vector of C.

Use point A $(1, 2, 2)$ and the direction vector $\vec{AC} = 4\vec{i} + 2\vec{j} - \vec{k}$ to form the cartesian equation of the line.

Calculate the vector $\vec{AB}$ .

Calculate the cross product $\vec{AB} \times \vec{AC}$ .

Calculate the magnitude of $\vec{AC}$ and $\vec{AB} \times \vec{AC}$ .

Calculate the perpendicular distance $d$ from point B to line AC using the formula $d = \frac{|\vec{AB} \times \vec{AC}|}{|\vec{AC}|}$ .

Calculate the area of triangle ABC using the formula Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}|$ .

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More Mathematics Questions

Triangle ABC has vertices with position vectors i+2j+2k, 2i+5j-2k and 5i+4j+k. Find (a) AC, (b) the cartesian equation of line AC, (c) (i) the perpendicular distance of point B from line AC, (ii) the area of triangle ABC.

Step-by-step solution

Calculate the vector AC⃗\vec{AC}AC by subtracting the position vector of A from the position vector of C.

Use point A (1,2,2)(1, 2, 2)(1,2,2) and the direction vector AC⃗=4i⃗+2j⃗−k⃗\vec{AC} = 4\vec{i} + 2\vec{j} - \vec{k}AC=4i+2j​−k to form the cartesian equation of the line.

Calculate the vector AB⃗\vec{AB}AB.

Calculate the cross product AB⃗×AC⃗\vec{AB} \times \vec{AC}AB×AC.

Calculate the magnitude of AC⃗\vec{AC}AC and AB⃗×AC⃗\vec{AB} \times \vec{AC}AB×AC.

Calculate the perpendicular distance ddd from point B to line AC using the formula d=∣AB⃗×AC⃗∣∣AC⃗∣d = \frac{|\vec{AB} \times \vec{AC}|}{|\vec{AC}|}d=∣AC∣∣AB×AC∣​.

Calculate the area of triangle ABC using the formula Area =12∣AB⃗×AC⃗∣= \frac{1}{2} |\vec{AB} \times \vec{AC}|=21​∣AB×AC∣.