If a times r = b + lambda a and a cdot r = 3, | vedclass

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(B) Given $a 	imes r = b + \lambda a$ and $a \cdot r = 3.$Taking the dot product with $a$ on both sides of $a 	imes r = b + \lambda a$:$(a 	imes r)

Vedclass · Accepted Answer

$r = \frac{7}{6}i + \frac{2}{3}j, \lambda = \frac{5}{6}$

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(B) Given $a 	imes r = b + \lambda a$ and $a \cdot r = 3.$Taking the dot product with $a$ on both sides of $a 	imes r = b + \lambda a$:$(a 	imes r) \cdot a = b \cdot a + \lambda (a \cdot a)$Since $(a 	imes r) \cdot a = 0$,we have $0 = b \cdot a + \lambda |a|^2$.$a = 2i + j - k \implies |a|^2 = 2^2 + 1^2 + (-1)^2 = 6$.$b \cdot a = (-i - 2j + k) \cdot (2i + j - k) = -2 - 2 - 1 = -5$.Thus,$0 = -5 + \lambda(6) \implies \lambda = \frac{5}{6}$.Now,$a 	imes r = b + \frac{5}{6}a$.Taking the cross product with $a$ on both sides:$a 	imes (a 	imes r) = a 	imes (b + \frac{5}{6}a) = a 	imes b + 0$.Using the vector triple product formula $a 	imes (a 	imes r) = (a \cdot r)a - (a \cdot a)r$:$(3)a - 6r = a 	imes b$.$a 	imes b = \begin{vmatrix} i & j & k \ 2 & 1 & -1 \ -1 & -2 & 1 \end{vmatrix} = i(1-2) - j(2-1) + k(-4+1) = -i - j - 3k$.$3(2i + j - k) - 6r = -i - j - 3k$.$6i + 3j - 3k - 6r = -i - j - 3k$.$6r = 7i + 4j \implies r = \frac{7}{6}i + \frac{2}{3}j$.

If $a \times r = b + \lambda a$ and $a \cdot r = 3,$ where $a = 2i + j - k$ and $b = -i - 2j + k,$ then $r$ and $\lambda$ are equal to

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