Download App

VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA1 Mark
Question
The vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ satisfy the equation $2\vec{\text{a}}+\vec{\text{b}}=\vec{\text{p}}$ and $\vec{\text{a}}+2\vec{\text{b}}=\vec{\text{q}},$ where $\vec{\text{p}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{q}}=\hat{\text{i}}-\hat{\text{j}}.$ If $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}},$ then:
  1. $\cos \theta = \frac{4}{5}$
  2. $\sin \theta = \frac{1}{\sqrt{2}}$
  3. $\cos \theta = -\frac{4}{5}$
  4. $\cos \theta = -\frac{3}{5}$
✓

Answer

  1. $\cos \theta = -\frac{4}{5}$
Solution:
Given that
$2\vec{\text{a}}+\vec{\text{b}}=\vec{\text{p}}\dots(1)$
$\vec{\text{a}}+2\vec{\text{b}}=\vec{\text{q}}\dots(2)$
Solving these two we get
$\vec{\text{a}}=\frac{2\vec{\text{p}}-\vec{\text{q}}}{3},\vec{\text{b}}=\frac{2\vec{\text{q}}-\vec{\text{p}}}{3}$
And we have
$\vec{\text{p}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{q}}=\hat{\text{i}}-\hat{\text{j}}$
Substituting the values of $\vec{\text{p}}$ and $\vec{\text{q}},$ we get
$\vec{\text{a}}=\frac{2\vec{\text{p}}-\vec{\text{q}}}{3}=\frac{2\big(\hat{\text{i}}+\hat{\text{j}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)}{3}=\frac{\hat{\text{i}}+3\hat{\text{j}}}{3}$
$\Rightarrow|\vec{\text{a}}|=\frac{1}{3}\sqrt{1+9}=\frac{\sqrt{10}}{3}$
$\vec{\text{b}}=\frac{2\vec{\text{q}}-\vec{\text{p}}}{3}=\frac{2\big(\hat{\text{i}}-\hat{\text{j}}\big)-\big(\hat{\text{i}}+\hat{\text{j}}\big)}{3}=\frac{\hat{\text{i}}-3\hat{\text{j}}}{3}$
$\Rightarrow\big|\vec{\text{b}}\big|=\frac{1}{3}\sqrt{1+9}=\frac{\sqrt{10}}{3}$
$\vec{\text{a}}.\vec{\text{b}}=\frac{1}{9}(1-9)=\frac{-8}{9}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\frac{-8}{9}=\frac{\sqrt{10}}{3}\times\frac{\sqrt{10}}{3}\cos\theta$
$\Rightarrow\frac{-8}{9}=\frac{10}{9}\cos\theta$
$\Rightarrow\cos\theta=\frac{-8}{9}\times\frac{9}{10}=\frac{-4}{5}$

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

Similar questions

The corner points of the feasible region determined by the system of linear inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If the maximum value of z = ax + by, where a, b > 0 occurs at both (2, 4) and (4, 0), then:
  1. a = 2b
  2. 2a = b
  3. a = b
  4. 3a = b
→
What is the magnitude of vector -3i + 5j:
  1. $\sqrt{34}$
  2. $\sqrt{32}$
  3. $\sqrt{8}$
  4. $\sqrt{16}$
→
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}-\text{Ky}=0, \text{y}(0)=1$ approaches to zero when $\text{x}\rightarrow\propto$ if,
  1. K = 0
  2. K > 0
  3. K < 0
  4. None of these.
→
If the diraction ratios of a line are proportional to 1, -3, 2, then its diraction cosines are:
  1. $\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
  2. $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
  3. $-\frac{1}{\sqrt{14}},\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
  4. $-\frac{1}{\sqrt{14}},-\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}$
→
If $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$ Then f(x) is continuous at $\text{x}=\frac{\pi}{2},$ if:
  1. $\text{a}=\frac{1}{3},\text{ b}=2$
  2. $\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
  3. $\text{a}=\frac{2}{3},\text{ b}=\frac{8}{3}$
  4. none of these
→
The direction ratios of the line perprndicular to the lines $\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$ and, $\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$ are proportional to:
  1. 4, 5, 7
  2. 4, -5, 7
  3. 4, -5, -7
  4. -4, 5, 7
→
A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. Then, domain of R is:
  1. {2, 3, 5}
  2. {3, 5}
  3. {2, 3, 4}
  4. {2, 3, 4, 5}
→
If A is a singular matrix, then adj A is:
  1. Non-singular.
  2. Singular.
  3. Symmetric.
  4. Not defined.
→
A vector parallel to the line of intersection of the plance $\vec{\text{r}}.(3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=1$ and $\vec{\text{r}}.(\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}})=2$ is:
→
The area bounded by curve $y = x^2- 1$ and tangents to it at $(2, 3)$ and $y -$ axis is:
→