The unit normal vector to the line joining i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the points be $A(1, -1)$ and $B(2, 3)$.The vector along the line $AB$ is $\vec{AB} = (2-1)i + (3 - (-1))j = i + 4j$.$A$ vector perpendicular t

Vedclass · Accepted Answer

$\frac{-4i + j}{\sqrt{17}}$

Vedclass · Answer

(B) Let the points be $A(1, -1)$ and $B(2, 3)$.The vector along the line $AB$ is $\vec{AB} = (2-1)i + (3 - (-1))j = i + 4j$.$A$ vector perpendicular to $\vec{AB} = i + 4j$ is of the form $\lambda(4i - j)$.The unit normal vector is $\pm \frac{4i - j}{\sqrt{4^2 + (-1)^2}} = \pm \frac{4i - j}{\sqrt{17}}$.The line equation passing through $(1, -1)$ with direction vector $i + 4j$ is $\frac{x-1}{1} = \frac{y+1}{4}$,which simplifies to $4x - y - 5 = 0$.The normal vector to the line $ax + by + c = 0$ is $ai + bj$. Here,the normal vector is $4i - j$.To check the direction towards the origin $(0, 0)$,we test the point $(0, 0)$ in the expression $4x - y - 5$. Since $4(0) - 0 - 5 = -5 Thus,the required unit normal vector is $-\frac{4i - j}{\sqrt{17}} = \frac{-4i + j}{\sqrt{17}}$.

The unit normal vector to the line joining $i - j$ and $2i + 3j$ and pointing towards the origin is

Similar Questions

The coordinates of the foot of the perpendicular drawn from the point $(-2, 3)$ to the line $3x - y - 1 = 0$ are:

If $(h, k)$ is the image of the point $(2, -3)$ with respect to the line $5x - 3y = 2$,then $h + k =$

If $(h, k)$ is the image of the point $(3, -4)$ with respect to the line $2x - 3y - 5 = 0$ and $(\ell, m)$ is the foot of the perpendicular from $(h, k)$ onto the line $3x + 2y + 12 = 0$,then $\ell h + mk + 1 =$ ?

The perpendicular bisector of the line segment joining $P(1, 4)$ and $Q(k, 3)$ has a $y-$intercept of $-4$. Then a possible value of $k$ is:

The equation of the line perpendicular to the line $ax + by + c = 0$ and passing through $(a, b)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module