We are given:<\/p>\n\n\n\n
- \n
- A vector u<\/strong> = (5, -12)<\/li>\n\n\n\n
- A scalar c<\/strong> = -3<\/li>\n\n\n\n
- We are asked to find the magnitude<\/strong> of the vector cu<\/strong>, which is written as \u2225cu\u2225\\| c\\mathbf{u} \\|.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep-by-Step Solution:<\/h3>\n\n\n\n
1. Understand scalar multiplication:<\/strong><\/h4>\n\n\n\n
When we multiply a vector u<\/strong> by a scalar c<\/strong>, we multiply each component<\/strong> of the vector by the scalar: cu=c\u22c5(5,\u221212)=(\u22123)(5,\u221212)=(\u221215,36)c\\mathbf{u} = c \\cdot (5, -12) = (-3)(5, -12) = (-15, 36)<\/p>\n\n\n\n
\n\n\n\n2. Find the magnitude of the new vector \u2225cu\u2225\\| cu \\|:<\/strong><\/h4>\n\n\n\n
The magnitude of a 2D vector (x,y)(x, y) is given by: \u2225(x,y)\u2225=x2+y2\\| (x, y) \\| = \\sqrt{x^2 + y^2}<\/p>\n\n\n\n
So for the vector (\u221215,36)(-15, 36): \u2225(\u221215,36)\u2225=(\u221215)2+(36)2=225+1296=1521=39\\| (-15, 36) \\| = \\sqrt{(-15)^2 + (36)^2} = \\sqrt{225 + 1296} = \\sqrt{1521} = 39<\/p>\n\n\n\n
\n\n\n\nFinal Answer: B \u2013 39<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation <\/h3>\n\n\n\n
The magnitude of a vector represents its length or size regardless of direction. Given a vector u = (5, -12)<\/strong> and a scalar c = -3<\/strong>, we are to find the magnitude of the scaled vector cu<\/strong>.<\/p>\n\n\n\n
To begin, scalar multiplication involves multiplying each component of the vector by the scalar. Here, multiplying each component of u<\/strong> by -3<\/strong> gives us a new vector cu = (-15, 36)<\/strong>. Notice that this changes the direction of the vector (because the scalar is negative), but not its magnitude in absolute terms.<\/p>\n\n\n\n
Next, we find the magnitude of the new vector. The formula for the magnitude of a 2D vector (x, y) is: \u2225(x,y)\u2225=x2+y2\\| (x, y) \\| = \\sqrt{x^2 + y^2}<\/p>\n\n\n\n
Applying this to (-15, 36)<\/strong>: \u2225(\u221215,36)\u2225=(\u221215)2+362=225+1296=1521=39\\| (-15, 36) \\| = \\sqrt{(-15)^2 + 36^2} = \\sqrt{225 + 1296} = \\sqrt{1521} = 39<\/p>\n\n\n\n
The magnitude remains the same as if we scaled the original magnitude directly. Indeed, an alternative method confirms this: \u2225cu\u2225=\u2223c\u2223\u22c5\u2225u\u2225\\| c\\mathbf{u} \\| = |c| \\cdot \\| \\mathbf{u} \\| \u2225u\u2225=52+(\u221212)2=25+144=169=13\\| \\mathbf{u} \\| = \\sqrt{5^2 + (-12)^2} = \\sqrt{25 + 144} = \\sqrt{169} = 13<\/p>\n\n\n\n
So: \u2225\u22123\u22c5u\u2225=\u2223\u22123\u2223\u22c513=3\u22c513=39\\| -3 \\cdot \\mathbf{u} \\| = |-3| \\cdot 13 = 3 \\cdot 13 = 39<\/p>\n\n\n\n
Thus, the correct answer is B \u2013 39<\/strong>.<\/p>\n\n\n\n
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- A scalar c<\/strong> = -3<\/li>\n\n\n\n