If (alpha, beta, gamma) is a triad of real nu | vedclass

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(B) Given the vector equation: $\hat{i}-2 \hat{j}+5 \hat{k}=\alpha(\hat{i}+\hat{j}+\hat{k})+\beta(\hat{i}+2 \hat{j}+3 \hat{k})+\gamma(2 \hat{i}-\hat{j

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) Given the vector equation: $\hat{i}-2 \hat{j}+5 \hat{k}=\alpha(\hat{i}+\hat{j}+\hat{k})+\beta(\hat{i}+2 \hat{j}+3 \hat{k})+\gamma(2 \hat{i}-\hat{j}+\hat{k})$.Comparing the coefficients of $\hat{i}, \hat{j}, 	ext{ and } \hat{k}$ on both sides, we get the system of linear equations:$1 = \alpha + \beta + 2\gamma$  $(1)$$-2 = \alpha + 2\beta - \gamma$  $(2)$$5 = \alpha + 3\beta + \gamma$  $(3)$Adding $(2)$ and $(3)$: $3 = 2\alpha + 5\beta \implies 2\alpha + 5\beta = 3$  $(4)$Subtracting $(2)$ from $(3)$: $7 = \beta + 2\gamma \implies \beta + 2\gamma = 7$  $(5)$From $(1)$, $\alpha + \beta + 2\gamma = 1$. Substituting $(5)$ into this: $\alpha + 7 = 1 \implies \alpha = -6$.Substitute $\alpha = -6$ into $(4)$: $2(-6) + 5\beta = 3 \implies -12 + 5\beta = 3 \implies 5\beta = 15 \implies \beta = 3$.Substitute $\beta = 3$ into $(5)$: $3 + 2\gamma = 7 \implies 2\gamma = 4 \implies \gamma = 2$.Now, calculate $\alpha^2 - \beta^2 + \gamma^2 = (-6)^2 - (3)^2 + (2)^2 = 36 - 9 + 4 = 31$.

If $(\alpha, \beta, \gamma)$ is a triad of real numbers satisfying $\hat{i}-2 \hat{j}+5 \hat{k}=\alpha(\hat{i}+\hat{j}+\hat{k})+\beta(\hat{i}+2 \hat{j}+3 \hat{k})+\gamma(2 \hat{i}-\hat{j}+\hat{k}),$ then $\alpha^2-\beta^2+\gamma^2=$

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