Let ABC be a triangle such that overrightarro | vedclass

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

Question

(D) In a triangle $ABC$,$\overrightarrow{BC} + \overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{0}$,so $\overrightarrow{a} + \overrightarro

Vedclass · Accepted Answer

Both $(S1)$ and $(S2)$ are false.

Vedclass · Answer

(D) In a triangle $ABC$,$\overrightarrow{BC} + \overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{0}$,so $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}$.From $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}$,we have $\overrightarrow{b} + \overrightarrow{c} = -\overrightarrow{a}$.Squaring both sides: $|\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{b} \cdot \overrightarrow{c}) = |\overrightarrow{a}|^2$.Given $|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$,we have $(2\sqrt{3})^2 + |\overrightarrow{c}|^2 + 2(12) = (6\sqrt{2})^2$.$12 + |\overrightarrow{c}|^2 + 24 = 72 \implies |\overrightarrow{c}|^2 = 36 \implies |\overrightarrow{c}| = 6$.For $(S1)$: $|(\overrightarrow{a} 	imes \overrightarrow{b}) + (\overrightarrow{c} 	imes \overrightarrow{b})| - |\overrightarrow{c}| = |(\overrightarrow{a} + \overrightarrow{c}) 	imes \overrightarrow{b}| - |\overrightarrow{c}|$.Since $\overrightarrow{a} + \overrightarrow{c} = -\overrightarrow{b}$,this becomes $|(-\overrightarrow{b}) 	imes \overrightarrow{b}| - |\overrightarrow{c}| = |\overrightarrow{0}| - 6 = -6$.Thus,$(S1)$ is false.For $(S2)$: $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0} \implies \overrightarrow{a} + \overrightarrow{c} = -\overrightarrow{b}$.$|\overrightarrow{a} + \overrightarrow{c}|^2 = |-\overrightarrow{b}|^2 \implies |\overrightarrow{a}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{c}) = |\overrightarrow{b}|^2$.$72 + 36 + 2(\overrightarrow{a} \cdot \overrightarrow{c}) = 12 \implies 2(\overrightarrow{a} \cdot \overrightarrow{c}) = -96 \implies \overrightarrow{a} \cdot \overrightarrow{c} = -48$.Using $\cos(\angle ABC) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} = \frac{(-\overrightarrow{c}) \cdot \overrightarrow{a}}{|-\overrightarrow{c}| |\overrightarrow{a}|} = \frac{-(\overrightarrow{a} \cdot \overrightarrow{c})}{6 \cdot 6\sqrt{2}} = \frac{48}{36\sqrt{2}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.Since $\frac{2\sqrt{2}}{3} 
eq \sqrt{\frac{2}{3}}$,$(S2)$ is false.

Similar Questions

The scalar product of vectors $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$ and a unit vector along the sum of vectors $\bar{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\bar{c}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}$ is $1$. Then the value of $\lambda$ is:

If $a$ and $b$ are unit vectors and $a - b$ is also a unit vector,then the angle between $a$ and $b$ is

Let $\bar{a}$ and $\bar{b}$ be two non-collinear unit vectors. If $\bar{u}=\bar{a}-(\bar{a} \cdot \bar{b}) \bar{b}$ and $\bar{v}=\bar{a} \times \bar{b}$,then $|\bar{v}|=$

If $A = (k, 1, -1)$,$B = (2k, 0, 2)$,and $C = (2 + 2k, k, 1)$,and $AB \perp BC$,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module