6 Show that the lines with equations r=left(begin{array}{l}1 1 2end{array}right)+lambdaleft(begin{array}{l}1 0 3end{array}right), mathbf{r}=left(begin{array}{c}-2 3 -1end{array}right)+muleft(begin{array}{c}1 -1 0end{array}right) and mathbf{r}=left(begin{array}{c}2 -1 -1end{array}right)+tleft(begin{array}{c}-1 2 3end{array}right) form a triangle, and find its area.

Question

6 Show that the lines with equations r=left(begin{array}{l}1  1  2end{array}right)+lambdaleft(begin{array}{l}1  0  3end{array}right), mathbf{r}=left(begin{array}{c}-2  3  -1end{array}right)+muleft(begin{array}{c}1  -1  0end{array}right) and mathbf{r}=left(begin{array}{c}2  -1  -1end{array}right)+tleft(begin{array}{c}-1  2  3end{array}right) form a triangle, and find its area.

Accepted Answer

Step:1 ：First, we write the parametric equations of the lines as:(mathbf{r}_1 = begin{pmatrix} 1+lambda  1  2+3lambda end{pmatrix}), (mathbf{r}_2 = begin{pmatrix} -2+mu  3-mu  -1 end{pmatrix}), and (mathbf{r}_3 = begin{pmatrix} 2-t  -1+2t  -1+3t end{pmatrix})Step:2 ：Next, we find the intersection points of the lines by setting the corresponding components equal. For (mathbf{r}_1) and (mathbf{r}_2), we have (1+lambda = -2+mu) and (2+3lambda = -1). Solving for (lambda) and (mu), we get (lambda = -1) and (mu = 0). Thus, the intersection point is (A = begin{pmatrix} 0  1  -1 end{pmatrix})Step:3 ：Similarly, for (mathbf{r}_2) and (mathbf{r}_3), we have (-2+mu = 2-t) and (3-mu = -1+2t). Solving for (t) and (mu), we get (t = 1) and (mu = 2). Thus, the intersection point is (B = begin{pmatrix} 0  1  1 end{pmatrix})Step:4 ：Finally, for (mathbf{r}_1) and (mathbf{r}_3), we have (1+lambda = 2-t) and (2+3lambda = -1+3t). Solving for (t) and (lambda), we get (t = 2) and (lambda = 0). Thus, the intersection point is (C = begin{pmatrix} 1  1  2 end{pmatrix})Step:5 ：Now that we have the vertices of the triangle, we can find the side lengths. We have (AB = sqrt{(0-0)^2+(1-1)^2+(-1-1)^2} = sqrt{4} = 2), (BC = sqrt{(1-0)^2+(1-1)^2+(2-1)^2} = sqrt{2}), and (CA = sqrt{(1-0)^2+(1-1)^2+(2+1)^2} = sqrt{10})Step:6 ：Using Heron&#x27;s formula, we first find the semi-perimeter: (s = frac{2+sqrt{2}+sqrt{10}}{2})Step:7 ：Then, we find the area of the triangle: (A = sqrt{s(s-2)(s-sqrt{2})(s-sqrt{10})})Step:8 ：Calculating the area, we get (A = sqrt{frac{1}{2}(2+sqrt{2}+sqrt{10})(sqrt{2}+sqrt{10})(sqrt{10}-2)(sqrt{2}-2)})Step:9 ：Finally, we simplify the expression to get the area of the triangle: (A = boxed{sqrt{2}})

Answer

Step: 1 ：First, we write the parametric equations of the lines as:(mathbf{r}_1 = begin{pmatrix} 1+lambda  1  2+3lambda end{pmatrix}), (mathbf{r}_2 = begin{pmatrix} -2+mu  3-mu  -1 end{pmatrix}), and (mathbf{r}_3 = begin{pmatrix} 2-t  -1+2t  -1+3t end{pmatrix})Step: 2 ：Next, we find the intersection points of the lines by setting the corresponding components equal. For (mathbf{r}_1) and (mathbf{r}_2), we have (1+lambda = -2+mu) and (2+3lambda = -1). Solving for (lambda) and (mu), we get (lambda = -1) and (mu = 0). Thus, the intersection point is (A = begin{pmatrix} 0  1  -1 end{pmatrix})Step: 3 ：Similarly, for (mathbf{r}_2) and (mathbf{r}_3), we have (-2+mu = 2-t) and (3-mu = -1+2t). Solving for (t) and (mu), we get (t = 1) and (mu = 2). Thus, the intersection point is (B = begin{pmatrix} 0  1  1 end{pmatrix})Step: 4 ：Finally, for (mathbf{r}_1) and (mathbf{r}_3), we have (1+lambda = 2-t) and (2+3lambda = -1+3t). Solving for (t) and (lambda), we get (t = 2) and (lambda = 0). Thus, the intersection point is (C = begin{pmatrix} 1  1  2 end{pmatrix})Step: 5 ：Now that we have the vertices of the triangle, we can find the side lengths. We have (AB = sqrt{(0-0)^2+(1-1)^2+(-1-1)^2} = sqrt{4} = 2), (BC = sqrt{(1-0)^2+(1-1)^2+(2-1)^2} = sqrt{2}), and (CA = sqrt{(1-0)^2+(1-1)^2+(2+1)^2} = sqrt{10})Step: 6 ：Using Heron&#x27;s formula, we first find the semi-perimeter: (s = frac{2+sqrt{2}+sqrt{10}}{2})Step: 7 ：Then, we find the area of the triangle: (A = sqrt{s(s-2)(s-sqrt{2})(s-sqrt{10})})Step: 8 ：Calculating the area, we get (A = sqrt{frac{1}{2}(2+sqrt{2}+sqrt{10})(sqrt{2}+sqrt{10})(sqrt{10}-2)(sqrt{2}-2)})Step: 9 ：Finally, we simplify the expression to get the area of the triangle: (A = boxed{sqrt{2}})

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