Question:

Step 1: Calculate $a \times b$ and its magnitude. Given vectors are $a = 2i + j − 2k$ and $b = i + j$. The cross product $a \times b$ is calculated as: $$a \times b = \begin{vmatrix} i & j & k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = i(1 \cdot 0 - (-2) \cdot 1) - j(2 \cdot 0 - (-2) \cdot 1) + k(2 \cdot 1 - 1 \cdot 1)$$ $$a \times b = i(0 + 2) - j(0 + 2) + k(2 - 1) = 2i - 2j + k$$ The magnitude of $a \times b$ is: $$|a \times b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$ Step 2: Determine the magnitude of vector $c$. The problem states that $|(a \times b) \times c| = 3$ and the angle between $c$ and $a \times b$ is $30°$. The magnitude of the cross product of two vectors $u$ and $v$ is given by $|u \times v| = |u||v|\sin(\theta)$, where $\theta$ is the angle between $u$ and $v$. Let $u = a \times b$ and $v = c$. $$|(a \times b) \times c| = |a \times b||c|\sin(30°)$$ Substitute the known values: $$3 = (3)|c|\left(\frac{1}{2}\right)$$ $$3 = \frac{3}{2}|c|$$ $$|c| = 2$$ Step 3: Calculate $a \cdot c$ using the condition $|c − a| = 3$. The magnitude of vector $a$ is: $$|a| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ Given $|c − a| = 3$, we can square both sides: $$|c − a|^2 = 3^2$$ $$(c − a) \cdot (c − a) = 9$$ Expand the dot product: $$|c|^2 - 2(a \cdot c) + |a|^2 = 9$$ Substitute the magnitudes $|c|=2$ and $|a|=3$: $$2^2 - 2(a \cdot c) + 3^2 = 9$$ $$4 - 2(a \cdot c) + 9 = 9$$ $$13 - 2(a \cdot c) = 9$$ $$2(a \cdot c) = 13 - 9$$ $$2(a \cdot c) = 4$$ $$a \cdot c = 2$$