If a=2 hat{i}-3 hat{j}+hat{k}, b=hat{i}-hat{j | vedclass

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(C) Given vectors are $a=2 \hat{i}-3 \hat{j}+\hat{k}$, $b=\hat{i}-\hat{j}+2 \hat{k}$, and $c=2 \hat{i}+\hat{j}+\hat{k}$.Using the vector triple produc

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$\sqrt{\frac{11}{39}}|a 	imes(b 	imes c)|$

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(C) Given vectors are $a=2 \hat{i}-3 \hat{j}+\hat{k}$, $b=\hat{i}-\hat{j}+2 \hat{k}$, and $c=2 \hat{i}+\hat{j}+\hat{k}$.Using the vector triple product formula $(a 	imes b) 	imes c = (a \cdot c)b - (b \cdot c)a$:$a \cdot c = (2)(2) + (-3)(1) + (1)(1) = 4 - 3 + 1 = 2$$b \cdot c = (1)(2) + (-1)(1) + (2)(1) = 2 - 1 + 2 = 3$$(a 	imes b) 	imes c = 2(i - j + 2k) - 3(2i - 3j + k) = (2i - 2j + 4k) - (6i - 9j + 3k) = -4i + 7j + k$$|(a 	imes b) 	imes c| = \sqrt{(-4)^2 + 7^2 + 1^2} = \sqrt{16 + 49 + 1} = \sqrt{66}$.Now, for $a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$:$a \cdot c = 2$$a \cdot b = (2)(1) + (-3)(-1) + (1)(2) = 2 + 3 + 2 = 7$$a 	imes (b 	imes c) = 2(i - j + 2k) - 7(2i + j + k) = (2i - 2j + 4k) - (14i + 7j + 7k) = -12i - 9j - 3k$$|a 	imes (b 	imes c)| = \sqrt{(-12)^2 + (-9)^2 + (-3)^2} = \sqrt{144 + 81 + 9} = \sqrt{234}$.Thus, $\frac{|(a 	imes b) 	imes c|}{|a 	imes (b 	imes c)|} = \sqrt{\frac{66}{234}} = \sqrt{\frac{11}{39}}$.Therefore, $|(a 	imes b) 	imes c| = \sqrt{\frac{11}{39}}|a 	imes (b 	imes c)|$.

If $a=2 \hat{i}-3 \hat{j}+\hat{k}$, $b=\hat{i}-\hat{j}+2 \hat{k}$ and $c=2 \hat{i}+\hat{j}+\hat{k}$ are three vectors, then $|(a \times b) \times c|=$

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