Download App

Scalar Or Dot Product — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsScalar Or Dot Product5 Marks
Question
If $\big|\vec{\text{a}}+\vec{\text{b}}\big|=60,\big|\vec{\text{a}}-\vec{\text{b}}\big|=40$ and $\big|\vec{\text{b}}\big|=46,$ find $|\vec{\text{a}}|$
✓

Answer

Here, $\big|\vec{\text{a}}+\vec{\text{b}}\big|=60$
Squaring both the sides,
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=(60)^2$
$(\vec{\text{a}}+\vec{\text{b}})=(60)^2$
$(\vec{\text{a}})^2+(\vec{\text{b}})^2+2\vec{\text{a}}\vec{\text{b}}=3600$
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}\vec{\text{b}}=3600\dots(1)$
Now, $\big|\vec{\text{a}}-\vec{\text{b}}\big|=40$
Squaring both the sides,
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=(40)^2$
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}\vec{\text{b}}=1600\dots(2)$
Adding (1) and (2),
$2|\vec{\text{a}}|^2+2\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}\vec{\text{b}}-2\vec{\text{a}}\vec{\text{b}}=3600-1600$
$2|\vec{\text{a}}|^2+2(46)^2=5200$
$2|\vec{\text{a}}|^2=5200-4232$
$2|\vec{\text{a}}|^2=968$
$|\vec{\text{a}}|^2=\frac{968}{2}$
$|\vec{\text{a}}|^2=484$
$|\vec{\text{a}}|=\sqrt{484}$
$|\vec{\text{a}}|=22$

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 "Solve the Following Question.(5 Marks)" from Scalar Or Dot Product→Scalar Or Dot Product — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $\sec ^{-1}\left(\frac{7 x^3-5 y^3}{7 x^3+5 y^3}\right)=m$, show that $\frac{d^2 y}{d x^2}=0$
→
Integrate the following integrals:
$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
→
Prove that $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.
→
The equation of the tangent at $(2,3)$ on the curve $y^2=a x^3+b$ is $y=4 x-5$. Find the values of $a$ and $b$.
→
If $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x},$ find $\frac{\text{dy}}{\text{dx}}$
→
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta),\text{y}=\text{ae}^\theta(\sin\theta+\cos\theta)$
→
Find the cartesian form of the equations of the following planes.
$\vec{\text{r}}=(1+\text{s}+\text{t})\hat{\text{i}}+(2-\text{s}+\text{t})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
→
Let O be the origin. We define a relation between two points P and Q in a plane if OP = OQ. Show that the relation, so defined is an equivalence relation.
→
Maximum Z = 5x + 3y Subject to $2\text{x}+\text{y}\geq10$ $\text{x}+3\text{y}\geq15$ $\text{x}\leq10$$\text{y}\leq8$
$\text{x},\text{y}\geq0$
→
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\sec^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big),$ if:
$\text{x}\in\Big(0,\frac{1}{\sqrt{2}}\Big)$
→