The equation of a line in Cartesian form pass | vedclass

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(B) Step $1$: Calculate the cross product $v = a 	imes b = (2i + j + 2k) 	imes (3i - 4j)$.$v = \begin{vmatrix} i & j & k \ 2 & 1 & 2 \ 3 & -4 & 0

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$x/8 = y/6 = z/-11$

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(B) Step $1$: Calculate the cross product $v = a 	imes b = (2i + j + 2k) 	imes (3i - 4j)$.$v = \begin{vmatrix} i & j & k \ 2 & 1 & 2 \ 3 & -4 & 0 \end{vmatrix} = i(0 - (-8)) - j(0 - 6) + k(-8 - 3) = 8i + 6j - 11k$.Step $2$: The line passes through $(0, 0, 0)$ and $(4, 3, c)$. The direction vector of the line is proportional to $(4, 3, c)$. Since the line is parallel to $v = (8, 6, -11)$, we have $(4, 3, c) = k(8, 6, -11)$.Step $3$: Comparing components, $4 = 8k \implies k = 1/2$. Then $3 = 6k$ (consistent) and $c = -11k = -11(1/2) = -5.5$.Step $4$: The line passes through $(0, 0, 0)$ with direction ratios $(8, 6, -11)$. The Cartesian equation is $x/8 = y/6 = z/-11$.

The equation of a line in Cartesian form passing through $(0, 0, 0)$ and $(4, 3, c)$ and parallel to $a \times b$ where $a = 2i + j + 2k$ and $b = 3i - 4j$ is given by the direction vector $a \times b$. Find the value of $c$ and the equation of the line.

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