Q.Find the direction cosines of the straight line passing through the points $(5,6,7)$ and $(7,9,13)$. Also, find the parametric form of vector equation and Cartesian equations of the straight line passing through two given points.

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 $t=1+2(-1)=-1$. …