Download App

VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA5 Marks
Question
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two non-collinear vectors having the same initial point. What are the vectors represented by $\vec{\text{a}}+\vec{\text{b}}\text{ and }\vec{\text{a}}-\vec{\text{b}}$.
✓

Answer

Here, it is given that $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two non-collinear vectors having the same initial point.Let $\vec{\text{a}}=\overrightarrow{\text{AB}}\text{ and }\vec{\text{b}}=\overrightarrow{\text{AD}}$, So we can draw a parallelogram ABCD as above.
By the properties of parallelogram
$\overrightarrow{\text{BC}}=\vec{\text{b}}\text{ and }\overrightarrow{\text{DC}}=\vec{\text{a}}$
In $\triangle\text{ABC}$,
Using triangle law,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
$\vec{\text{a}}+\vec{\text{b}}=\overrightarrow{\text{AC}}\ \dots(\text{i})$
In $\triangle\text{ABD}$
$\overrightarrow{\text{AD}}+\overrightarrow{\text{DB}}=\overrightarrow{\text{AB}}$
$\vec{\text{b}}+\overrightarrow{\text{DB}}=\vec{\text{a}}$
$\overrightarrow{\text{DB}}=\vec{\text{a}}-\vec{\text{b}}\ \dots(\text{ii})$
From (i) and (ii), we get that
$\vec{\text{a}}+\vec{\text{b}}\text{ and }\vec{\text{a}}-\vec{\text{b}}$ are diagonals of a parallelogram whose adjacent sides are $\vec{\text{a}}\text{ and }\vec{\text{b}}$

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 "5 Marks Questions" from VECTOR ALGEBRA→VECTOR ALGEBRA — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Evaluate the following intregals:
$\int\frac{1}{\text{x}\log\text{x}(2+\log\text{x})}\text{ dx}$
→
Show that the semi-vertical angle of the cone of the maximum value and of given slant height is ${\tan ^{ - 1}}\sqrt 2$ 
→
Find the distance of the point (1, -2, 3) from the plane x - y + z = 5 measured along a line parallel to $\frac{\text{x}}{2}=\frac{\text{y}}{3}=\frac{\text{z}}{-6}.$
→
Find the intervals in which the following functions are increasing or decreasing.
$f(x) = 3x^4 - 4x^3- 12x^2 + 5$
→
If $\tan(\text{x}+\text{y})+\tan(\text{x}+\text{y})=1,$ find $\frac{\text{dy}}{\text{dx}}$
→
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2\sqrt{2}}}\Big)$
→
Find the image of the point $(0,2,3)$ in the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$
→
Differentiate $\cos^{-1}(4\text{x}^3-3\text{x})$ with respect to $\tan^{-1}\Big(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\Big),$ if $\frac{1}{2}<\text{x}<1$
→
A furniture trader deals in only two items – chairs and tables. He has ₹ 50,000 to invest and a space to store at most 35 items. A chair costs him ₹ 1000 and a table costs him ₹ 2000 The trader earns a profit of ₹ 150 and ₹ 250 on a chair and table, respectively. Formulate the above problem as an LPP to maximise the profit and solve it graphically.
→
Solve the following differential equation:
$(\sin\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=2\sin^2\text{x}\cos\text{x}$
→