1) __ Solved Problem #3 3a. Sketch the vector v=3i-3j and find its magnitude. __ . Pencil Problem #36 3a. Sketch the vector v=3i+j and find its magnitude.

To sketch the vector \( v = 3i - 3j \), we can plot it on a coordinate plane. The vector starts at the origin (0,0) and ends at the point (3,-3).<br /><br />To find the magnitude of the vector \( v = 3i - 3j \), we can use the formula for the magnitude of a vector in two dimensions:<br /><br />\[ \|v\| = \sqrt{x^2 + y^2} \]<br /><br />where \( x \) and \( y \) are the components of the vector.<br /><br />In this case, \( x = 3 \) and \( y = -3 \). Plugging these values into the formula, we get:<br /><br />\[ \|v\| = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]<br /><br />Therefore, the magnitude of the vector \( v = 3i - 3j \) is \( 3\sqrt{2} \).<br /><br />To sketch the vector \( v = 3i + j \), we can plot it on a coordinate plane. The vector starts at the origin (0,0) and ends at the point (3,1).<br /><br />To find the magnitude of the vector \( v = 3i + j \), we can use the same formula for the magnitude of a vector in two dimensions:<br /><br />\[ \|v\| = \sqrt{x^2 + y^2} \]<br /><br />In this case, \( x = 3 \) and \( y = 1 \). Plugging these values into the formula, we get:<br /><br />\[ \|v\| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]<br /><br />Therefore, the magnitude of the vector \( v = 3i + j \) is \( \sqrt{10} \).