If overline{a}=2 hat{imath}-hat{jmath}+hat{k} | vedclass

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(C) Since the given vectors $\overline{a}, \overline{b},$ and $\overline{c}$ are coplanar,their scalar triple product must be zero,i.e.,$[\overline{a}

Vedclass · Accepted Answer

$x^{2}+3 x=4$

Vedclass · Answer

(C) Since the given vectors $\overline{a}, \overline{b},$ and $\overline{c}$ are coplanar,their scalar triple product must be zero,i.e.,$[\overline{a} \overline{b} \overline{c}] = 0$.This is equivalent to the determinant of the matrix formed by their components being zero:$\begin{vmatrix} 2 & -1 & 1 \ 1 & 2 & -3 \ 3 & \lambda & 5 \end{vmatrix} = 0$Expanding the determinant along the first row:$2(2 	imes 5 - (-3) 	imes \lambda) - (-1)(1 	imes 5 - (-3) 	imes 3) + 1(1 	imes \lambda - 2 	imes 3) = 0$$2(10 + 3\lambda) + 1(5 + 9) + 1(\lambda - 6) = 0$$20 + 6\lambda + 14 + \lambda - 6 = 0$$7\lambda + 28 = 0$$7\lambda = -28 \Rightarrow \lambda = -4$Now,we check which equation has $x = -4$ as a root:$(A)$ $(-4)^2 + 2(-4) = 16 - 8 = 8 
eq 6$$(B)$ $(-4)^2 + 2(-4) = 16 - 8 = 8 
eq 4$$(C)$ $(-4)^2 + 3(-4) = 16 - 12 = 4$. This is correct.$(D)$ $(-4)^2 + 3(-4) = 16 - 12 = 4 
eq 6$Thus,$\lambda = -4$ is the root of the equation $x^2 + 3x = 4$.

If $\overline{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}$,$\overline{b}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k}$ and $\overline{c}=3 \hat{\imath}+\lambda \hat{\jmath}+5 \hat{k}$ are coplanar,then $\lambda$ is the root of the equation

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