Respuesta :
Answer:
The required equations are
[tex](-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0[/tex] and
[tex](-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0[/tex].
Step-by-step explanation:
The given parametric equation of the line, [tex]L[/tex], is [tex]x=5+t, y=6, z=-2-3t,[/tex] so, an arbitrary point on the line is [tex]R(x,y,z)=R(5+t, 6, -2-3t)[/tex]
The vector equation of the line passing through the points [tex]P(-5,7,-8)[/tex] and [tex]R(5+t, 6, -2-3t)[/tex] is
[tex]\vec P + \lambda \vec{(PR)}=0[/tex]
[tex]\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((5+t-(-5))\hat i + (6-7)\hat j +(-2-3t-8)\hat k\right)=0[/tex]
[tex]\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+t)\hat i -\hat j +(6-3t)\hat k\right)=0\;\cdots (i)[/tex]
The given equation can also be written as
[tex]\frac {x-5}{1}=\frac {v-6}{0}=\frac{z+2}{-3}=t \; \cdots (ii)[/tex]
The standard equation of the line passes through the point [tex]P_0(x_0,y_0,z_0)[/tex] and having direction[tex]\vec v= a_1 \hat i +a_2 \hat j +a_3 \hat k[/tex] is
[tex]\frac {x-x_0}{a_1}=\frac {y-y_0}{a_2}=\frac{z-z_0}{a_3}=t \;\cdots (iii)[/tex]
Here, The value of the parameter,[tex]t[/tex], of any point [tex]R[/tex] at a distance [tex]d[/tex] from the point, [tex]P_0[/tex], can be determined by
[tex]|t \vec v|=d\;\cdots (iv)[/tex]
Comparing equations [tex](ii)[/tex] and [tex](iii)[/tex]
The line is passing through the point [tex]P_0 (5,6,-2)[/tex] having direction [tex]\vec v=\hat i -3 \hat k[/tex].
Note that the point [tex]Q(5,6,-2)[/tex] is the same as [tex]P_0[/tex] obtained above.
Now, the value of the parameter, [tex]t[/tex], for point [tex]R[/tex] at distance [tex]d=3[/tex] from the point [tex]Q(5,6,-2)[/tex] can be determined by equation [tex](iv)[/tex], we have
[tex]|t(\hat i -3 \hat k)|=3[/tex]
[tex]\Rightarrow t^2|(\hat i -3 \hat k)|^2=9[/tex]
[tex]\Rightarrow 10t^2=9[/tex]
[tex]\Rightarrow t^2=\frac {9}{10}[/tex]
[tex]\Rightarrow t=\pm \frac {3}{\sqrt {10}}[/tex]
Put the value of [tex]t[/tex] in equation [tex](i)[/tex], the required equations are as follows:
For [tex]t= \frac {3}{\sqrt {10}}[/tex]
[tex](-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +\left(6-3\times \frac {3}{\sqrt {10}})\hat k\right)=0[/tex]
[tex]\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0[/tex]
and for [tex]t= -\frac {3}{\sqrt {10}}[/tex],
[tex](-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\left (-\frac {3}{\sqrt {10}}\right))\hat i -\hat j +(6-3\times \left(-\frac {3}{\sqrt {10}}\right)\hat k\right)=0[/tex]
[tex]\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0[/tex]
