If the vertices A, B, C of a triangle ABC are | vedclass

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

Question

(A) The vertices of $	riangle ABC$ are $A(1,2,3), B(-1,0,0),$ and $C(0,1,2)$.$\angle ABC$ is the angle between the vectors $\overrightarrow{BA}$ and

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{10}{\sqrt{102}}\right)$

Vedclass · Answer

(A) The vertices of $	riangle ABC$ are $A(1,2,3), B(-1,0,0),$ and $C(0,1,2)$.$\angle ABC$ is the angle between the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$.$\overrightarrow{BA} = (1 - (-1))\hat{i} + (2 - 0)\hat{j} + (3 - 0)\hat{k} = 2\hat{i} + 2\hat{j} + 3\hat{k}$.$\overrightarrow{BC} = (0 - (-1))\hat{i} + (1 - 0)\hat{j} + (2 - 0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k}$.$\overrightarrow{BA} \cdot \overrightarrow{BC} = (2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10$.$|\overrightarrow{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.$|\overrightarrow{BC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.Using the dot product formula: $\cos(\angle ABC) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|} = \frac{10}{\sqrt{17} 	imes \sqrt{6}} = \frac{10}{\sqrt{102}}$.Therefore,$\angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}ight)$.

If the vertices $A, B, C$ of a triangle $ABC$ are $(1,2,3), (-1,0,0), (0,1,2)$ respectively,then find $\angle ABC$. $[\angle ABC \text{ is the angle between the vectors } \overrightarrow{BA} \text{ and } \overrightarrow{BC}]$.

Similar Questions

Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{C}$. Then $\overrightarrow{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$ is :

Let $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be two sides of a triangle. The median $\overrightarrow{AM}$ is perpendicular to the angle bisector $\overrightarrow{OL}$ and $|\overrightarrow{AM}|:|\overrightarrow{OL}|=1:2$. The angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is

If $a \cdot i = a \cdot (i + j) = a \cdot (i + j + k)$, then $a = $

If $\vec{l}, \vec{m}, \vec{n}$ are coplanar vectors,for what value of $\lambda$ are the points with position vectors $\vec{l} - 2\vec{m} + 3\vec{n}$,$2\vec{l} + \lambda\vec{m} - 4\vec{n}$,and $-7\vec{m} + 10\vec{n}$ collinear?

If $a(\vec{\alpha} \times \vec{\beta}) + b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \overrightarrow{0}$, where $a, b, c$ are non-zero scalars, then the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module