The point(s) on the line vec r = hat i + hat | vedclass

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(D) The equation of the line is given by $\vec r = (1+t)\hat i + (1+3t)\hat j + (1-t)\hat k$. Any point on this line is of the form $P(1+t, 1+3t, 1-t)

Vedclass · Accepted Answer

$(\frac{7}{5}, \frac{11}{5}, \frac{3}{5}), (- \frac{11}{5}, - \frac{43}{5}, \frac{21}{5})$

Vedclass · Answer

(D) The equation of the line is given by $\vec r = (1+t)\hat i + (1+3t)\hat j + (1-t)\hat k$. Any point on this line is of the form $P(1+t, 1+3t, 1-t)$. The equation of the plane is $x + 2y + 2z + 2 = 0$. The distance $d$ of point $P$ from the plane is given by $d = \frac{|(1+t) + 2(1+3t) + 2(1-t) + 2|}{\sqrt{1^2 + 2^2 + 2^2}} = 3$. Simplifying the numerator: $|1 + t + 2 + 6t + 2 - 2t + 2| = |5t + 7|$. So,$\frac{|5t + 7|}{3} = 3$,which implies $|5t + 7| = 9$. Case $1$: $5t + 7 = 9 \Rightarrow 5t = 2 \Rightarrow t = \frac{2}{5}$. Substituting $t = \frac{2}{5}$ into the point coordinates: $x = 1 + \frac{2}{5} = \frac{7}{5}$,$y = 1 + 3(\frac{2}{5}) = \frac{11}{5}$,$z = 1 - \frac{2}{5} = \frac{3}{5}$. Point is $(\frac{7}{5}, \frac{11}{5}, \frac{3}{5})$. Case $2$: $5t + 7 = -9 \Rightarrow 5t = -16 \Rightarrow t = -\frac{16}{5}$. Substituting $t = -\frac{16}{5}$ into the point coordinates: $x = 1 - \frac{16}{5} = -\frac{11}{5}$,$y = 1 + 3(-\frac{16}{5}) = -\frac{43}{5}$,$z = 1 - (-\frac{16}{5}) = \frac{21}{5}$. Point is $(-\frac{11}{5}, -\frac{43}{5}, \frac{21}{5})$. Thus,the points are $(\frac{7}{5}, \frac{11}{5}, \frac{3}{5})$ and $(-\frac{11}{5}, -\frac{43}{5}, \frac{21}{5})$.

The point$(s)$ on the line $\vec r = \hat i + \hat j + \hat k + t(\hat i + 3\hat j - \hat k)$ at a distance of $3 \ units$ from the plane $\vec r \cdot (\hat i + 2\hat j + 2\hat k) + 2 = 0$ are

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