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irga5000 [103]
3 years ago
7

By vector methods, find the cosine of the angle between the lines (x - 1)/(3) = (y - 0.5)/(2) = z and x = y = z

Mathematics
1 answer:
melamori03 [73]3 years ago
3 0

Answer:

The angle between the lines \frac{x-1}{3}= \frac{y-0.5}{2}=\frac{z-0}{1} and \frac{x-0}{1}= \frac{y-0}{1}=\frac{z-0}{1} is \sqrt{\frac{6}{7}}

Step-by-step explanation:

The equation of a line with direction vector \vec{d}=(l,m.n) that passes through the point (x_{1},y_{1},z_{1}) is given by the formula

\frac{x-x_{1}}{l}= \frac{y-x_{1}}{m}=\frac{z-z_{1}}{n}, where <em>l</em>,<em>m</em>, and <em>n</em> are non-zero real numbers.

This is called the <u><em>symmetric equations of the line</em></u>.

The angle between two lines \frac{x-x_{1}}{l_{1} }= \frac{y-y_{1}}{m_{1} }=\frac{z-z_{1}}{n_{1}} and \frac{x-x_{2}}{l_{2} }= \frac{y-y_{2}}{m_{2} }=\frac{z-z_{2}}{n_{2}} equal the angle subtended by direction vectors, d_{1} and d_{2} of the lines

cos (\theta)=\frac{\vec{d_{1}}\cdot\vec{d_{2}}}{|\vec{d_{1}}|\cdot|\vec{d_{2}}|}=\frac{l_{1} \cdot\l_{2}+m_{1} \cdot\ m_{2}+n_{1} \cdot\ n_{2}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}} \cdot \sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}

Given that

\frac{x-1}{3}= \frac{y-0.5}{2}=\frac{z-0}{1} and \frac{x-0}{1}= \frac{y-0}{1}=\frac{z-0}{1}

l_{1}=3, m_{1}=2,n_{1}=1\\ l_{2}=1, m_{2}=1,n_{2}=1

We can use the formula above to find the cosine of the angle between the lines

cos(\theta)=\frac{3 \cdot 1+2 \cdot 1 +1 \cdot 1}{\sqrt{3^{2}+2^{2}+1^{2}} \cdot \sqrt{1^{2}+1^{2}+1^{2}}} = \sqrt{\frac{6}{7}}

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