If A(-1, 2, 3),B(1, 1, 1) and C(2, -1, 3) are | vedclass

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

Question

(A) Given points are $A(-1, 2, 3)$,$B(1, 1, 1)$,and $C(2, -1, 3)$.First,we find two vectors lying on the plane:$\overrightarrow{AB} = (1 - (-1))i + (1

Vedclass · Accepted Answer

$\pm \left( \frac{2i + 2j + k}{3} \right)$

Vedclass · Answer

(A) Given points are $A(-1, 2, 3)$,$B(1, 1, 1)$,and $C(2, -1, 3)$.First,we find two vectors lying on the plane:$\overrightarrow{AB} = (1 - (-1))i + (1 - 2)j + (1 - 3)k = 2i - j - 2k$$\overrightarrow{AC} = (2 - (-1))i + (-1 - 2)j + (3 - 3)k = 3i - 3j + 0k$To find a normal vector to the plane,we calculate the cross product $\overrightarrow{AB} 	imes \overrightarrow{AC}$:$\overrightarrow{AB} 	imes \overrightarrow{AC} = \begin{vmatrix} i & j & k \ 2 & -1 & -2 \ 3 & -3 & 0 \end{vmatrix} = i(0 - 6) - j(0 - (-6)) + k(-6 - (-3)) = -6i - 6j - 3k$Let $\vec{n} = -6i - 6j - 3k$. The magnitude is $|\vec{n}| = \sqrt{(-6)^2 + (-6)^2 + (-3)^2} = \sqrt{36 + 36 + 9} = \sqrt{81} = 9$.The unit normal vector is $\hat{n} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{-6i - 6j - 3k}{9} = \pm \left( \frac{-2i - 2j - k}{3} ight) = \pm \left( \frac{2i + 2j + k}{3} ight)$.

If $A(-1, 2, 3)$,$B(1, 1, 1)$ and $C(2, -1, 3)$ are points on a plane,then a unit normal vector to the plane $ABC$ is:

Similar Questions

If the length of the perpendicular drawn from the origin to a plane is $7$ units and its direction ratios are $-3, 2, 6$,then the equation of the plane is:

Consider the three planes $P_{1}: 3x + 15y + 21z = 9$; $P_{2}: x - 3y - z = 5$; and $P_{3}: 2x + 10y + 14z = 5$. Then,which one of the following is true?

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $2x + 3y - z = 5$.

If the planes $\bar{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3$ and $\bar{r} \cdot(4 \hat{i}-\hat{j}+\mu \hat{k})=5$ are parallel,then the values of $\lambda$ and $\mu$ are respectively:

$A$ plane $\pi$ passes through the points $(5,1,2)$, $(3,-4,6)$, and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosines of a normal to the plane $\pi$, then $|3l+2m+5n|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module