Download App

Algebra of Vectors — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSAlgebra of Vectors3 Marks
Question
ABCDE is a pentagon, prove that,$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
✓

Answer

Given ABCDE is a pentagon.
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$
$=\Big(\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\Big)+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$ $\Big[$Using triangle law in $\triangle\text{ABC},\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=\Big(\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}\Big)+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$
$=\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$ $\Big[$Using triangle law in $\triangle\text{ACD},\ \overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}\Big]$
$=\overrightarrow{\text{AD}}+\overrightarrow{\text{DA}}$
$=\overrightarrow{\text{AD}}-\Big(-\overrightarrow{\text{AD}}\Big)$
$=\vec0$
$\therefore\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "3 Marks Question" from Algebra of Vectors→Algebra of Vectors — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})=9$
→
Solve the following:
$\cos^{-1}\text{x}+\sin^{-1}\frac{\text{x}}{2}=\frac{\pi}{6}$
→
If the mapping f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g= {(2, 3), (5, 1), (1, 3)}, then write fog.
→
Using vectors show that the points A(-2, 3, 5), B(7, 0, -1), C(-3. -2, -5) and D(3, 4, 7) are such that AB and CD intersect at the point P(1, 2, 3).
→
Evaluate:
$\cos\Big(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)$
→
Differentiate the following w.r.t. x:
$2^{\cos^2}\text{x}$
→
Solve the following $\text{LPP}$ graphically:
Minimise $Z=5 x+10 y$
subject to the constraints $x+2 y \geq 120$
$x+y \geq 60, x-2 y \geq 0 \text { and } x, y \geq 0$
→
For what value of k is the following function continuous at x = 1
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1}, & \text{x} \neq 1\\\text{k}, & \text{x}= 1\end{cases}$ 
→
Assume that the chances of a patient having heart attack is $40\%.$ It is also assumed that meditation and yoga course reduces the risk of heart attack by $30\%$ and prescription of certain drug reduces its chances by $25\%.$  At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options and patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
→
Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
→