$\min .(|u||\vec{v} \times \vec{w}| \cos \theta)=-\alpha \sqrt{3401}$
$\cos \theta=-1$
$|u|=\alpha \text { (Given) }$
$|\vec{v} \times \vec{w}|=\sqrt{3401}$
$\vec{v} \times \vec{w}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2 \alpha & 1 & -1\end{array}\right|$
$\vec{v} \times \vec{w}=\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k}$
$|\vec{v} \times \vec{w}|=\sqrt{1+25 \alpha^2+9 \alpha^2}=\sqrt{3401}$
$34 \alpha^2=3400$
$\alpha^2=100$
$\alpha=10 \quad(\text { as } \alpha > 0)$
So $\vec{u} =\lambda(\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k})$
$\vec{u} =\sqrt{\lambda^2+25 \alpha^2 \lambda^2+9 \alpha^2 \lambda}$
$\alpha^2 =\lambda^2\left(1+25 \alpha^2+9 \alpha^2\right)$
$100 =\lambda^2(1+34 \times 100)$
$\lambda^2 =\frac{100}{3401}=\frac{m}{n}$
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$I.$ Adifferentiable function $' f '$ with maximum at $x = c$ ==> $ f "(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$. $\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx} $
$IV.$ If $f (x)$ has a maxima at $x = c$ , then $'f '$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \rightarrow 0$ for $h > 0.$ Now indicate the correct alternative.
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