Download App

VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA5 Marks
Question
Prove that a necessary and sufficient condition for three vectors $\vec{\text{a}},\ \vec{\text{b}}$ and $\vec{\text{c}}$ to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$.
✓

Answer

Necessary Condition: Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are three coplanar vectors. Then one of them can be expressed as the linear combination of other two vectors. Let, $\vec{\text{c}}=\text{x}\vec{\text{a}}+\text{y}\vec{\text{b}}$ $\text{x}\vec{\text{a}}+\text{y}\vec{\text{b}}-\vec{\text{c}}=0$ Put $\text{x}=1,\ \text{y}=\text{m},\ (-1)=\text{n}$$\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$
Thus, if $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors, then there exist scalars l, m, n $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ Such that l, m, n are not all zero simultaneously. Sufficient Condition: Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ be three vectors such that there exist scalars l, m, n not all zero simultaneously satisfying $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ $\text{n}\vec{\text{c}}=-\text{l}\vec{\text{a}}-\text{m}\vec{\text{b}}$ Dividing by n, both the sides $\frac{\text{n}\vec{\text{c}}}{\text{n}}=\frac{-\text{l}\vec{\text{a}}}{\text{n}}-\frac{\text{m}\vec{\text{b}}}{\text{n}}$ $\vec{\text{c}}=\Big(-\frac{\text{l}}{\text{n}}\Big)\vec{\text{a}}+\Big(-\frac{\text{m}}{\text{n}}\Big)\vec{\text{b}}$ $\vec{\text{c}}$ is a linear combination of $\vec{\text{a}} \text{ and }\vec{\text{b}}$ Hence, $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors.

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:
$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-1}\text{ dx}\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-1}\ \text{dx}$
→
If $\overrightarrow{\text{a}}\text{ and } \overrightarrow{\text{b}}$are two vectors such that $|\overrightarrow{\text{a}} + \overrightarrow{\text{b}}| = | \overrightarrow{\text{a}}|,$ then prove that vector 2 $\overrightarrow{\text{a}} + \overrightarrow{\text{b}}$is perpendicular to vector$\overrightarrow{\text{b}}.$
→
$\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)$
→
Evaluate: $\int\limits^{\pi/2}_{0}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\text{d}x$
→
Prove that the curves $\text{y}^{2} = 4x \text{ and } x^{2} = 4y $ divided the area of square bounded by $x = 0, x = 4,y=4 \text{ and } y = 0$ into three equal parts.
→
If $\text{x}=3\sin\text{t}-\sin3\text{t},$ $\text{y}=3\cos-\cos3\text{t}$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{t}=\frac{\pi}{3}.$
→
Solve the following differential equation:
$\text{dx + xdy}=\text{e}^{-\text{y}}\sec^2\text{y dy}$
→
Find the area of the ragion bounded by the curve $ay^{2 }= x^3 $ the $y-$ axis and the line $y = a$ and $y = 2a.$
→
Find the area of the region in the first quadrant enclosed by $x-$ axis, the line $\text{y}=\sqrt{3\text{x}}$ and the circle $x^2+ y^2 = 16.$
→
Solve the following differential equation: $(2\text{x}-10\text{y}^3)\frac{\text{dx}}{\text{dy}}+\text{y}=0$
→