The line passing through the point i + 3j + 2 | vedclass

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(D) The required line passes through the point $A = i + 3j + 2k$. Since the line is perpendicular to the lines with direction vectors $v_1 = 2i + j +

Vedclass · Accepted Answer

$r = i + 3j + 2k + \lambda (-i + 5j - 3k)$

Vedclass · Answer

(D) The required line passes through the point $A = i + 3j + 2k$. Since the line is perpendicular to the lines with direction vectors $v_1 = 2i + j + k$ and $v_2 = i + 2j + 3k$,its direction vector $b$ must be parallel to the cross product $v_1 	imes v_2$. Calculating the cross product: $b = v_1 	imes v_2 = \begin{vmatrix} i & j & k \ 2 & 1 & 1 \ 1 & 2 & 3 \end{vmatrix} = i(3 - 2) - j(6 - 1) + k(4 - 1) = i - 5j + 3k$. The equation of a line passing through point $a$ with direction vector $b$ is $r = a + \lambda b$. Thus,$r = (i + 3j + 2k) + \lambda (i - 5j + 3k)$. Note that the vector $(i - 5j + 3k)$ is parallel to $(-i + 5j - 3k)$. Therefore,$r = (i + 3j + 2k) + \lambda (-i + 5j - 3k)$ is also a valid representation of the same line.

The line passing through the point $i + 3j + 2k$ and perpendicular to the lines $r = (i + 2j - k) + \lambda (2i + j + k)$ and $r = (2i + 6j + k) + \mu (i + 2j + 3k)$ is:

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