Q.Find the points where the straight line passes through $(6,7,4)$ and $(8,4,9)$ cuts the $xz$ and $yz$ planes.

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 $t=1+2(-1)=-1$.
7. Check with the $x$-components: line 1 gives $x=-1-2(-1)=1$; line 2 gives $x=2+(-1)=1$ — consistent, so the lines do intersect.
8. Compute the point using either line: $x=1$, $y=2+t=2-1=1$, $z=3+t=3-1=2$.
9. Verify with line 2: $x=2+s=2-1=1$, $y=3+2s=3-2=1$, $z=5+3s=5-3=2$ — matches.

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.
4. The required line passes through $(1,3,5)$, i.e. has position vector $\vec i+3\vec j+5\vec k$, so its vector equation is $\vec r=(\vec i+3\vec j+5\vec k)+t\left(\vec i+5\vec j+\tfrac32\vec k\right)$.
5. This matches option (d); options (a) and (c) use the wrong point, and option (b) uses the un-normalised ratio $3$ instead of $3/2$ for the $z$-component.

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 $t=1$.
4. Test the same point against Line 2: from $x$: $-1+2s=0\Rightarrow s=0.5$. Check $y$: $-2+4(0.5)=0$ ✓. Check $z$: $-3-2(0.5)=-4$ ✓. So $(0,0,-4)$ also lies on Line 2 at $s=0.5$.
5. Since $(0,0,-4)$ lies on both lines for consistent parameter values, it is the common point of intersection.
6. (The other candidate points can be checked similarly and fail to satisfy both lines simultaneously.)
7. This matches option (a).

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 $t=1+2(-1)=-1$.
7. Verify with (i): LHS $=-1-2(-1)=-1+2=1$; RHS $=2+s=2+(-1)=1$. Consistent, confirming the lines genuinely intersect (not skew).
8. Substitute $t=-1$ into Line 1's point: $(-1-2(-1),\ 2+(-1),\ 3+(-1))=(1,\ 1,\ 2)$.
9. This matches option (c); the other options fail the consistency check across all three coordinate equations.