If a, b, c are three non-zero,non-coplanar ve | vedclass

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(B) The given vectors are constructed using the Gram-Schmidt orthogonalization process. For the set $\{a, b_1, c_2\}$: $1$. $a \cdot b_1 = a \cdot (b

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$\{a, b_1, c_2\}$

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(B) The given vectors are constructed using the Gram-Schmidt orthogonalization process. For the set $\{a, b_1, c_2\}$: $1$. $a \cdot b_1 = a \cdot (b - \frac{b \cdot a}{|a|^2} a) = a \cdot b - \frac{b \cdot a}{|a|^2} (a \cdot a) = a \cdot b - b \cdot a = 0$. Thus,$a \perp b_1$. $2$. $a \cdot c_2 = a \cdot (c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_1}{|b_1|^2} b_1) = a \cdot c - \frac{c \cdot a}{|a|^2} (a \cdot a) - \frac{c \cdot b_1}{|b_1|^2} (a \cdot b_1) = a \cdot c - c \cdot a - 0 = 0$. Thus,$a \perp c_2$. $3$. $b_1 \cdot c_2 = b_1 \cdot (c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_1}{|b_1|^2} b_1) = b_1 \cdot c - \frac{c \cdot a}{|a|^2} (b_1 \cdot a) - \frac{c \cdot b_1}{|b_1|^2} (b_1 \cdot b_1) = b_1 \cdot c - 0 - c \cdot b_1 = 0$. Thus,$b_1 \perp c_2$. Since all pairs are orthogonal,the set $\{a, b_1, c_2\}$ is a set of mutually orthogonal vectors. Therefore,option $(b)$ is the correct answer.

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