Question-1B Vector B has x y, and z components of 4.00,6.00, and 3.00 units , respectively. Calculate the magnitude of B and the angles B makes with the coordinate axes.

To calculate the magnitude of vector B, we can use the Pythagorean theorem. The magnitude of a vector is given by the square root of the sum of the squares of its components.<br /><br />Magnitude of B = √(x^2 + y^2 + z^2)<br /><br />Substituting the given values, we have:<br /><br />Magnitude of B = √(4.00^2 + 6.00^2 + 3.00^2)<br /><br />Magnitude of B = √(16 + 36 + 9)<br /><br />Magnitude of B = √61<br /><br />Therefore, the magnitude of vector B is √61 units.<br /><br />To find the angles that vector B makes with the coordinate axes, we can use the dot product formula. The dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them.<br /><br />For vector B and the x-axis:<br /><br />B · i = |B| * |i| * cos(θ)<br /><br />where B · i is the dot product of vector B and the unit vector i along the x-axis, |B| is the magnitude of vector B, |i| is the magnitude of the unit vector i, and θ is the angle between vector B and the x-axis.<br /><br />Since i is along the x-axis, its magnitude is 1. Therefore, the equation simplifies to:<br /><br />B · i = |B| * cos(θ)<br /><br />Substituting the given values, we have:<br /><br />4.00 = √61 * cos(θ)<br /><br />Solving for cos(θ), we get:<br /><br />cos(θ) = 4.00 / √61<br /><br />Therefore, the angle that vector B makes with the x-axis is given by:<br /><br />θ = arccos(4.00 / √61)<br /><br />Similarly, we can find the angles that vector B makes with the y-axis and z-axis by using the dot product formula and substituting the respective components of vector B.<br /><br />For vector B and the y-axis:<br /><br />B · j = |B| * |j| * cos(φ)<br /><br />where B · j is the dot product of vector B and the unit vector j along the y-axis, |B| is the magnitude of vector B, |j| is the magnitude of the unit vector j, and φ is the angle between vector B and the y-axis.<br /><br />Since j is along the y-axis, its magnitude is 1. Therefore, the equation simplifies to:<br /><br />B · j = |B| * cos(φ)<br /><br />Substituting the given values, we have:<br /><br />6.00 = √61 * cos(φ)<br /><br />Solving for cos(φ), we get:<br /><br />cos(φ) = 6.00 / √61<br /><br />Therefore, the angle that vector B makes with the y-axis is given by:<br /><br />φ = arccos(6.00 / √61)<br /><br />For vector B and the z-axis:<br /><br />B · k = |B| * |k| * cos(ψ)<br /><br />where B · k is the dot product of vector B and the unit vector k along the z-axis, |B| is the magnitude of vector B, |k| is the magnitude of the unit vector k, and ψ is the angle between vector B and the z-axis.<br /><br />Since k is along the z-axis, its magnitude is 1. Therefore, the equation simplifies to:<br /><br />B · k = |B| * cos(ψ)<br /><br />Substituting the given values, we have:<br /><br />3.00 = √61 * cos(ψ)<br /><br />Solving for cos(ψ), we get:<br /><br />cos(ψ) = 3.00 / √61<br /><br />Therefore, the angle that vector B makes with the z-axis is given by:<br /><br />ψ = arccos(3.00 / √61)<br /><br />In summary, the magnitude of vector B is √61 units, and the angles that vector B makes with the coordinate axes are given by:<br /><br />θ = arccos(4.00 / √61)<br /><br />φ = arccos(6.00 / √61)<br /><br />ψ = arccos(3.00 / √61)