Question

Let \((\overset{̄}{p}∧\overset{̄}{q})\) denote the angle between \(\overset{̄}{p}\) and \(\overset{̄}{q}\). If \(\overset{̄}{a}+\overset{̄}{b}+\overset{̄}{c} = \overset{̄}{0},|\overset{̄}{a}| = 7,|\overset{̄}{b}| = 5\) and \(|\overset{̄}{c}| = 3\) then (take \(π = \frac{22}{7}\))

• A. \(sin(\overset{̄}{b}∧\overset{̄}{c}) = \frac{1}{2}\)
• B. \(cos(\overset{̄}{a}∧\overset{̄}{c}) = -\frac{π}{4}\)
• C. \(cos(\overset{̄}{b}∧\overset{̄}{c}) = -\frac{1}{2}\)
• D. \(sin(\overset{̄}{a}∧\overset{̄}{c}) = \frac{π}{4}\)

Hint:

Square b = -(a + c) to find a dot c.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
If \(\vec a+\vec b+\vec c=\vec0\), the three vectors form a triangle. We can find angles between pairs by squaring one vector in terms of the other two.

Step 2: Angle between a and c:
\(\vec b=-(\vec a+\vec c)\). So \(|\vec b|^2=|\vec a|^2+|\vec c|^2+2\vec a\cdot\vec c\):
\[ 25=49+9+2\vec a\cdot\vec c\ \Rightarrow\ \vec a\cdot\vec c=-\frac{33}2 \]

Step 3: Cosine:
\[ \cos(\vec a\wedge\vec c)=\frac{-33/2}{7\times3}=-\frac{33}{42}=-\frac{11}{14} \]

Step 4: Check option (B):
With \(\pi=\dfrac{22}7\): \(-\dfrac\pi4=-\dfrac{22}{28}=-\dfrac{11}{14}\). This matches, so (B) is true.

Step 5: Check the others:
For \(\vec b\) and \(\vec c\): \(\vec a=-(\vec b+\vec c)\) gives \(49=25+9+2\vec b\cdot\vec c\), so \(\vec b\cdot\vec c=\tfrac{15}2\) and \(\cos(\vec b\wedge\vec c)=\tfrac12\). So (C) with \(-\tfrac12\) is false and (A) with \(\sin=\tfrac12\) is false, since \(\sin=\tfrac{\sqrt3}2\). For (D), \(\sin(\vec a\wedge\vec c)=\sqrt{1-\tfrac{121}{196}}=\tfrac{5\sqrt3}{14}\), not \(\tfrac{11}{14}\). So (D) is false.

Final Answer:
cos(a,c) = -11/14 = -pi/4 with pi = 22/7. \[ \boxed{\cos(\vec a\wedge\vec c)=-\frac\pi4} \]