Q.The vertices of $\triangle ABC$ are $A(7,2,1),\ B(6,0,3)$, and $C(4,2,4)$. Find $\angle ABC$.

Solving the parametric equations of the two given lines simultaneously gives the intersection point $(1,1,2)$.

1. Line 1: $\vec r=(-\vec i+2\vec j+3\vec k)+t(-2\vec i+\vec j+\vec k)$, i.e. $(x,y,z)=(-1-2t,\ 2+t,\ 3+t)$.
2. Line 2: $\vec r=(2\vec i+3\vec j+5\vec k)+s(\vec i+2\vec j+3\vec k)$, i.e. $(x,y,z)=(2+s,\ 3+2s,\ 5+3s)$.
3. Equate the $y$-components: $2+t=3+2s \Rightarrow t=1+2s$.
4. Equate the $z$-components: $3+t=5+3s \Rightarrow t=2+3s$.
5. Set the two expressions for $t$ equal: $1+2s=2+3s \Rightarrow -1=s \Rightarrow s=-1$.
6. Then $t=1+2(-1)=-1$. …

Rewriting the third ratio in standard form gives direction ratios $(1,5,3/2)$; the required line through $(1,3,5)$ is $\vec r=(\vec i+3\vec j+5\vec k)+t\left(\vec i+5\vec j+\tfrac32\vec k\right)$.

1. The given line is $\dfrac{x-3}{1}=\dfrac{y+3}{5}=\dfrac{2z-5}{3}$. Rewrite the third fraction so the coefficient of $z$ is $1$: $\dfrac{2z-5}{3}=\dfrac{2\left(z-\tfrac52\right)}{3}=\dfrac{z-\tfrac52}{\tfrac32}$.
2. So the direction ratios of the given line are $\left(1,\,5,\,\tfrac32\right)$.
3. A line parallel to it shares the same direction ratios. …

Write both lines in parametric form and check each candidate point against both sets of parametric equations; $(0,0,-4)$ satisfies both lines (at $t=1$ on line 1 and $s=0.5$ on line 2), confirming it as the intersection.

1. Line 1: $\dfrac{x-6}{-6}=\dfrac{y+4}{4}=\dfrac{z-4}{-8}=t$, giving parametric form $x=6-6t,\ y=-4+4t,\ z=4-8t$.
2. Line 2: $\dfrac{x+1}{2}=\dfrac{y+2}{4}=\dfrac{z+3}{-2}=s$, giving parametric form $x=-1+2s,\ y=-2+4s,\ z=-3-2s$.
3. Test option (a), $(0,0,-4)$, against Line 1: from $x$: $6-6t=0\Rightarrow t=1$. Check $y$: $-4+4(1)=0$ ✓. Check $z$: $4-8(1)=-4$ ✓. So $(0,0,-4)$ lies on Line 1 at $t=1$. …

Solving the three simultaneous parametric equations for $t$ and $s$ (with a consistency check) locates the intersection at $(1,1,2)$.

1. Line 1: $\vec r=(-1,2,3)+t(-2,1,1)$, so a general point is $(-1-2t,\ 2+t,\ 3+t)$.
2. Line 2: $\vec r=(2,3,5)+s(1,2,3)$, so a general point is $(2+s,\ 3+2s,\ 5+3s)$.
3. Equate coordinates:
   • $x$: $-1-2t=2+s$ … (i)
   • $y$: $2+t=3+2s$ … (ii)
   • $z$: $3+t=5+3s$ … (iii)
4. From (ii): $t=1+2s$. From (iii): $t=2+3s$.
5. Equate these two expressions for $t$: $1+2s=2+3s\Rightarrow -1=s\Rightarrow s=-1$.
6. Then $t=1+2(-1)=-1$. …