मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

Two Vectors Have Magnitudes 2 M and 3m. the Angle Between Them is 60°. Find (A) the Scalar Product of the Two Vectors, (B) the Magnitude of Their Vector Product.

Advertisements
Advertisements

प्रश्न

Two vectors have magnitudes 2 m and 3m. The angle between them is 60°. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.

थोडक्यात उत्तर
Advertisements

उत्तर

Let the two vectors be \[\left| \vec{a} \right| = 2 m\text { and } \left| \vec{b} \right| = 3 m\].

Angle between the vectors, θ = 60°
(a) The scalar product of two vectors is given by \[\vec{a} . \vec{b} = \left| \vec{a} \right| . \left| \vec{b} \right| \cos\theta^\circ\]

∴ \[\vec{a} . \vec{b} = \left| \vec{a} \right| . \left| \vec{b} \right| \cos 60^\circ\]

\[= 2 \times 3 \times \frac{1}{2} = 3 m^2\]

(b) The vector product of two vectors is given by \[\left| \vec{a} \times \vec{b} \right| = \left| \vec{a} \right| \left| \vec{a} \right| \sin\theta^\circ\].

∴ \[\left| \vec{a} \times \vec{b} \right| = \left| \vec{a} \right| \left| \vec{a} \right| \sin 60^\circ\]

\[= 2 \times 3 \times \frac{\sqrt{3}}{2}\]

\[ = 3\sqrt{3} m^2\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Physics and Mathematics - Exercise [पृष्ठ २९]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 2 Physics and Mathematics
Exercise | Q 11 | पृष्ठ २९

संबंधित प्रश्‍न

What are the dimensions of volume of a sphere of radius a?


\[\int\frac{dx}{\sqrt{2ax - x^2}} = a^n \sin^{- 1} \left[ \frac{x}{a} - 1 \right]\] 
The value of n is


Find the dimensions of Planck's constant h from the equation E = hv where E is the energy and v is the frequency.


Find the dimensions of the coefficient of linear expansion α and


The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SI and CGS units using the following data : Specific gravity of mercury = \[13 \cdot 6\] , Density of \[\text{ water} = {10}^3 kg/ m^3 , g = 9 \cdot 8 m/ s^2\] at Calcutta. Pressure
= hpg in usual symbols.


Test if the following equation is dimensionally correct:
\[v = \frac{1}{2 \pi}\sqrt{\frac{mgl}{I}};\] 
where h = height, S = surface tension, \[\rho\] = density, P = pressure, V = volume, \[\eta =\] coefficient of viscosity, v = frequency and I = moment of interia.


Is a vector necessarily changed if it is rotated through an angle?


Is the vector sum of the unit vectors  \[\vec{i}\] and \[\vec{i}\] a unit vector? If no, can you multiply this sum by a scalar number to get a unit vector?

 


Let \[\vec{A} = 3 \vec{i} + 4 \vec{j}\]. Write a vector \[\vec{B}\] such that \[\vec{A} \neq \vec{B}\], but A = B.


If \[\vec{A} \times \vec{B} = 0\] can you say that

(a) \[\vec{A} = \vec{B} ,\]

(b) \[\vec{A} \neq \vec{B}\] ?


Let \[\vec{A} = 5 \vec{i} - 4 \vec{j} \text { and } \vec{B} = - 7 \cdot 5 \vec{i} + 6 \vec{j}\]. Do we have \[\vec{B} = k \vec{A}\] ? Can we say \[\frac{\vec{B}}{\vec{A}}\] = k ?


The component of a vector is 


Let \[\vec{A} \text { and } \vec{B}\] be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angle 30° and 60° respectively, find the resultant.


Refer to figure (2 − E1). Find (a) the magnitude, (b) x and y component and (c) the angle with the X-axis of the resultant of \[\overrightarrow{OA}, \overrightarrow{BC} \text { and } \overrightarrow{DE}\].


A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.


Suppose \[\vec{a}\] is a vector of magnitude 4.5 units due north. What is the vector (a) \[3 \vec{a}\], (b) \[- 4 \vec{a}\] ?


Prove that \[\vec{A} . \left( \vec{A} \times \vec{B} \right) = 0\].


Give an example for which \[\vec{A} \cdot \vec{B} = \vec{C} \cdot \vec{B} \text{ but } \vec{A} \neq \vec{C}\].


High speed moving particles are studied under


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×