Question

vector u has its initial point at (-7, 2) and its terminal point at (11, -5). vector v has a direction opposite that of vector u, and its magnitude is three times the magnitude of u. what is the component form of vector v?
a. ( mathbf{v} = langle -54, 63
angle )
b. ( mathbf{v} = langle -162, -63
angle )
c. ( mathbf{v} = langle -54, 21
angle )
d. ( mathbf{v} = langle -162, 21
angle )

Explanation:

Step1: Find component form of vector u

The component form of a vector with initial point \((x_1, y_1)\) and terminal point \((x_2, y_2)\) is \(\langle x_2 - x_1, y_2 - y_1
angle\). For vector \(u\), initial point \((-7, 2)\) and terminal point \((11, -5)\), so \(u=\langle11 - (-7), -5 - 2
angle=\langle18, - 7
angle\).

Step2: Determine vector v

Vector \(v\) has opposite direction of \(u\) and magnitude three times of \(u\). So \(v=- 3u\) (negative for opposite direction, 3 for three times magnitude). Multiply each component of \(u\) by \(-3\): \(v=\langle18\times(-3), - 7\times(-3)
angle=\langle - 54,21
angle\)? Wait, no, wait: Wait, opposite direction means scalar multiple is negative, and magnitude three times means scalar is \(- 3\)? Wait, no, wait: Wait, magnitude three times and opposite direction: the scalar is \(-3\)? Wait, no, let's recalculate. Wait, \(u=\langle18, - 7
angle\), opposite direction is \(-u=\langle - 18,7
angle\), three times magnitude is \(3\times(-u)=3\times\langle - 18,7
angle=\langle - 54,21
angle\)? No, wait, no: Wait, the magnitude of \(u\) is \(\sqrt{18^{2}+(-7)^{2}}\), magnitude of \(v\) is \(3\) times that, and direction opposite, so \(v = - 3u\)? Wait, no, if direction is opposite, the scalar is negative, and magnitude three times, so scalar is \(-3\)? Wait, \(u=\langle18, - 7
angle\), then \(v=-3\times\langle18, - 7
angle=\langle - 54,21
angle\)? But wait, let's check the options. Wait, option B is \(\langle - 162,-63
angle\), option A is \(\langle - 54,63
angle\), option C is \(\langle - 54,21
angle\), option D is \(\langle - 162,21
angle\). Wait, I think I made a mistake. Wait, no: Wait, initial point \((-7,2)\), terminal point \((11,-5)\). So \(x\)-component: \(11-(-7)=18\), \(y\)-component: \(-5 - 2=-7\), so \(u=\langle18, - 7
angle\). Then \(v\) is opposite direction (so scalar is negative) and magnitude three times, so \(v=-3u\)? Wait, no, opposite direction means the vector is \(-u\) (same magnitude, opposite direction), then three times magnitude means \(3\times(-u)=-3u\)? Wait, no, magnitude three times and opposite direction: the vector \(v\) is a scalar multiple of \(u\) with scalar \(k\) where \(|k| = 3\) and \(k<0\) (opposite direction), so \(k=-3\). So \(v=k\times u=-3\times\langle18, - 7
angle=\langle - 54,21
angle\)? But wait, let's check the options. Option C is \(\langle - 54,21
angle\), but wait, maybe I messed up the sign of the \(y\)-component. Wait, \(y\)-component of \(u\) is \(-5 - 2=-7\), so multiplying by \(-3\) gives \(-7\times(-3)=21\), yes. \(x\)-component: \(18\times(-3)=-54\). So \(v=\langle - 54,21
angle\), which is option C? Wait, but let's check again. Wait, maybe I made a mistake in the scalar. Wait, opposite direction: the vector \(v\) is parallel to \(-u\), and has magnitude three times that of \(u\). The magnitude of \(u\) is \(|u|=\sqrt{18^{2}+(-7)^{2}}=\sqrt{324 + 49}=\sqrt{373}\). The magnitude of \(v\) should be \(3\sqrt{373}\). The magnitude of \(\langle - 54,21
angle\) is \(\sqrt{(-54)^{2}+21^{2}}=\sqrt{2916 + 441}=\sqrt{3357}=3\sqrt{373}\) (since \(3357 = 9\times373\)), yes. So that's correct. Wait, but let's check the options again. Option C is \(\langle - 54,21
angle\), so that's the answer? Wait, but wait, maybe I messed up the direction. Wait, no, opposite direction means the vector is \(-u\) scaled by 3? Wait, no, \(v\) has direction opposite to \(u\), so \(v = k u\) where \(k<0\), and \(|v| = 3|u|\), so \(|k||u|=3|u|\), so \(|k| = 3\), so \(k=-3\), so \(v=-3u=\langle - 54,21
angle\), which is option C.

Answer:

C. \( \mathbf{v}=\langle -54, 21
angle \)