Consider a vector πβ, given as:
P = i+j
pxi + pyj = i+j
On comparing the components on both sides, we get:
Px = Py =1
|P| = \(\sqrt{P_x^2+P_y^2}\)= \(\sqrt{(1)^2+(1)^2}\)=\(\sqrt{2}\) ...(i)
Hence, the magnitude of the vector π β + πβ is \(\sqrt{2}\) .
Let π be the angle made by the vector πβ, with the x-axis, as shown in the following figure.
β΄tanΞΈ = \(\frac{P_y}{P_x}\)= 45 ...(ii)
Hence, the vector i + j makes an angle of 45 with the x- axis.
Let Q = i - j
Qx, i - Q , j = i-j
Qx = Qy = 1
|Q| = \(\sqrt{ Q_x^2 + Q_y^2}\) = \(\sqrt{2}\) ....(iii)
Hence, the magnitude of the vector π βββ πβ is \(\sqrt{2}\).
Let π be the angle made by the vector πβ, with the x-axis, as shown in the following figure.
β΄tanΞΈ = (\(\frac{Q_y}{Q_x}\))
ΞΈ = - tan-1 (\(\frac{-1}{-1}\)) = -45 ...(iv)
Hence, the vector i - j makes an angle of 45 with the x- axis.
It is given that:
A = 2i + 3j
ax i + Ay j = 2i + 3j
On comparing the coefficients of πβ and πβ, we have:
Ax = 2 and Ay = 3
|A| = β 22 + 33 = β13
Let π΄π₯ β make an angle π with the x-axis, as shown in the following figure.
β΄tanΞΈ = (Ay/Ax)
ΞΈ = - tan-1 (\(\frac{3}{2}\)) = tan-1 (1.5) = 56.31
Angle between the vectors (2i + 3j ) and (i + j). ΞΈ= 56.31 -45 = 11.31
Component of vector π΄β, along the direction of πββ, making and angle π.
= (AcosΞΈ) P = (Acos 11.31) \(\frac{(i+j)}{\sqrt2}\).
= \(\frac{\sqrt{13 \times 0.9806}}{ \sqrt{2 (i+j)}}\)
= 2.5 (i+j)
=\(\frac{25}{10}\) x \(\sqrt{2}\)
=\(\frac{5}{\sqrt2}\) ...(v)
Let ΞΈ be the angle between the vectors (2i + 3j ) and (i-j)
ΞΈ = 45 + 56.31 = 101.31
Component of vector π΄β, along the direction of πβ, making and angle π.
= (AcosΞΈ) Q = (AcosΞΈ) \(\frac{i-j}{\sqrt2}\)
= \(\sqrt{13}\) cos(901.31) \(\frac{(i-j)}{\sqrt2}\)
=\(\sqrt{\frac{13}{2}}\) sin 11.30 (i-j)
= -0.5 (i-j)
= \(\frac{-5}{10}\) x \(\sqrt{2}\)
=\(-\frac{1}{\sqrt2}\) ...(vi)
Car P is heading east with a speed V and car Q is heading north with a speed \(\sqrt{3}\). What is the velocity of car Q with respect to car P?
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
A physical quantity, represented both in magnitude and direction can be called a vector.
For the supplemental purposes of these vectors, there are two laws that are as follows;
It means that if we have any two vectors a and b, then for them
\(\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{b}+\overrightarrow{a}\)
It means that if we have any three vectors namely a, b and c.
\((\overrightarrow{a}+\overrightarrow{b})+\overrightarrow{c}=\overrightarrow{a}+(\overrightarrow{b}+\overrightarrow{c})\)
Read More: Addition of Vectors