Download App

VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA1 Mark
MCQ
If points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear, then a is equal to,
  • A
    40
  • ✓
    -40
  • C
    20
  • D
    -20
✓

Answer

Correct option: B.
-40
Given: Three points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear. Then,

$\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$

We have,

$\overrightarrow{\text{AB}}=\big(40\hat{\text{i}}-8\hat{\text{j}}\big)-\big(60\hat{\text{i}}+3\hat{\text{j}}\big)$

$=-20\hat{\text{i}}-11\hat{\text{j}}$

$\overrightarrow{\text{BC}}=\big(\text{a}\hat{\text{i}}-52\hat{\text{j}}\big)-\big(40\hat{\text{i}}-8\hat{\text{j}}\big)$

$=(\text{a}-40)\hat{\text{i}}-44\hat{\text{j}}$

Therefore,

$\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$

$\Rightarrow-20\hat{\text{i}}-11\hat{\text{j}}=\lambda(\text{a}-40)\hat{\text{i}}-\lambda44\hat{\text{j}}$

$\Rightarrow\lambda(\text{a}-40)=-20,\ -44\lambda=-11\Rightarrow\lambda=\frac{1}4$

$\Rightarrow\text{a}-40=-80$

$\Rightarrow\text{a}=-40$

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 "M.C.Q (1 Marks)" 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

The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is:
→
Choose the correct option from given four options$:\ \int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{dx}$ is equal to:
→
The function $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$ is continuous at $x = 0,$ then $k =$
→
Let $f$ be a real valued function defined by

$f(x) = sin^{-1} \left( {\frac{{\,\,1 - \,\,\left| x \right|}}{3}} \right) + cos^{-1}\left( {\frac{{\left| x \right|\,\, - \,\,3}}{5}} \right)$ .

Then domain of $f(x)$ is given by :

→
If $f(x) = cos \left[ {\frac{\pi }{x}} \right] cos \left( {\frac{\pi }{2}\,\,\left( {x\,\, - \,\,1} \right)} \right)$ then $f(x)$ is continuous at :

where $[x]$ is the greatest integerr function of $x$, 

→
Let $X$ denote the number of times heads occur in $n$ tosses of a fair coin. If $P(X = 4), P(X = 5)$ and $P(X = 6)$ are in $AP,$ the value of $n$ is:
→
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, then which of the following values of $\vec{\text{a}}.\vec{\text{b}}$ is not possible?
→
Let $f$ be an injective map with domain $\{x, y, z\}$ and range $\{1, 2, 3\}$, such that exactly one of the following statements is correct and the remaining are false.$\text{f(x)}=1,\ \text{f(y)}\neq1,\ \text{f(z)}\neq2.$The value of $f^{-1}(1)$ is:
→
The solution of differential equation $\frac{{dy}}{{dx}} + {\sin ^2}y = 0$ is
→
If $\int\text{f}(\text{x})\text{dx}=-2\cos\sqrt{\text{x}}+\text{c}$ then $f(x)$ is equal to:
→