Q.Find the vector equation of the line passing through the point $2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$ and parallel to the vector $4\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$.

The vector makes equal angles with the $x$ and $y$ axes and is perpendicular to the $z$ axis, so its $z$-component is zero and its $x$ and $y$ components are equal. With magnitude 3, each component is $\frac{3}{\sqrt{2}}$, giving the vector $\frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}$, which matches option (C) after rationalising.

The $x$-axis points along $\hat i=(1,0,0)$, so its direction ratios are $k,0,0$; answer (c).

In the vector form $\vec r = \vec a + m\vec b$, the coefficients of the parameter $m$ (i.e. $\vec b$) directly give the direction ratios of the line.

The line is $\vec r = 2\hat i - 3\hat j + \hat k + m(9\hat i - 2\hat j + 5\hat k)$.

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In the vector form $\vec r = \vec a + s\,\vec b$, the coefficients of the parameter give the line's direction ratios.

For $\vec r = 4\hat i - 7\hat j + 5\hat k + s(9\hat i - 2\hat j + 5\hat k)$, the fixed point is $(4,-7,5)$ and the direction vector is …

Direction ratios of a line joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are $(x_2-x_1,\,y_2-y_1,\,z_2-z_1)$.

Let the points be $P(2,3,-4)$ and $Q(1,3,-2)$.

Direction ratios of line PQ: …

For $ax+by+cz+d=0$, the normal has direction ratios $(a,b,c)=(3,4,5)$.

The normal to a plane $ax+by+cz+d=0$ has direction ratios equal to the coefficients $(a,b,c)$ …

Parallel lines have the same direction, so their direction ratios are proportional.

Direction ratios $(a,b,c)$ describe the direction a line points in. Two lines are parallel precisely when they point in the same (or exactly opposite) direction, which means one line's direction ratios are a scalar multiple of the oth …

Direction ratios $=(-1,-5,7)$ — option (iv).

Concept. For a line joining $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, the direction ratios are $x_2-x_1,\ y_2-y_1,\ z_2-z_1$.

Steps. With $A(2,3,-4)$ and $B(1,-2,3)$: …