Download App

VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA5 Marks
Question
If $\vec{\text{a}},\vec{\text{b}}$ are two non-collinear vectors, prove that the points with position vectors $\vec{\text{a}}+\vec{\text{b}},\ \vec{\text{a}}-\vec{\text{b}}$ and $\vec{\text{a}}+\lambda\vec{\text{b}}$ are collinear for all real values of $\lambda$.
✓

Answer

Let A, B, C be the points then, Position vector of $\text{A}=\vec{\text{a}}+\vec{\text{b}}$ Position vector of $\text{B}=\vec{\text{a}}-\vec{\text{b}}$ Position vector of $\text{C}=\vec{\text{a}}+\lambda\vec{\text{b}}$ $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A $=\big(\vec{\text{a}}-\vec{\text{b}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}\big)$ $=\vec{\text{a}}-\vec{\text{b}}-\vec{\text{a}}-\vec{\text{b}}$ $\overrightarrow{\text{AB}}=-2\vec{\text{b}}$ $\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B $=\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)-\big(\vec{\text{a}}-\vec{\text{b}}\big)$ $=\vec{\text{a}}+\lambda\vec{\text{b}}-\vec{\text{a}}+\vec{\text{b}}$ $\overrightarrow{\text{BC}}=(\lambda+1)\vec{\text{b}}$ Using $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$, we get $\overrightarrow{\text{AB}}=\Big[\frac{(\lambda+1)}{2}\Big]\Big(\overrightarrow{\text{BC}}\Big)$ Let $\Big(\frac{\lambda+1}{2}\Big)=\mu$Since $\lambda$ is a real number. So,
$\mu$ is also a real number. So, $\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$, but $\vec{\text{B}}$ is a common vector. Hence, A, B, C are collinear.

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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}}{\text{y + x}}$
→
Evaluate the integral $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$ using substitution.
→
$\begin{vmatrix}1&\text{a}&\text{a}^2\\\text{a}^2&1&\text{a}\\\text{a}&\text{a}^2&1\end{vmatrix}=(\text{a}^3-1)^2$
→
Find the particular solution of $\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1,$ that $\text{y}=3,$ when $\text{x}=0.$
→
If $\text{f(x)}=\frac{\tan\big(\frac{\pi}{4}-\text{x}\big)}{\cot2\text{x}}$ for $\text{x}\neq\frac{\pi}{4},$ find the value which can be assigned to f(x) at $\text{x}=\frac{\pi}{4}$ so that the function f(x) becomes continuous every where in $\Big[0,\frac{\pi}{2}\Big]$
→
Solve the following differential equation:
$\big(\text{y}^2-2\text{xy}\big)\text{dx}=\big(\text{x}^2-2\text{xy}\big)\text{dy}$
→
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.
→
Solve the following differential equation:
$(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x + 2y}$
→
The random variable $X$ can take only the values $0, 1, 2, 3.$ Given that $P(2) = P(3) = p$ and $P(0) = 2P(1)$. If $\Sigma p_ix_i^2 = 2\Sigma p_ix_i,$ find the value of $p.$
→
Find the shortest distance between the lines whose vector equations are:
$\vec{\text{r}}=(1-\text{t})\hat{\text{i}}+(\text{t}-2)\hat{\text{j}}+(3-2\text{t})\hat{\text{k}}\ \text{and}$
$\vec{\text{r}}=(\text{s}+1)\hat{\text{i}}+(2\text{s}-1)\hat{\text{j}}-(2\text{s}+1)\hat{\text{k}}$
→