Let vec{a} = 2hat{i} + hat{j} - hat{k} and ve | vedclass

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(A) Given $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.Then $\vec{a} + \vec{b} = 3\hat{i} + 3\hat{j} + 0\hat{

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) Given $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.Then $\vec{a} + \vec{b} = 3\hat{i} + 3\hat{j} + 0\hat{k} = 3(\hat{i} + \hat{j})$.$|\vec{a} + \vec{b}| = 3\sqrt{1^2 + 1^2} = 3\sqrt{2}$.Projection of $\vec{c}$ on $(\vec{a} + \vec{b})$ is $\frac{\vec{c} \cdot (\vec{a} + \vec{b})}{|\vec{a} + \vec{b}|} = 3\sqrt{2}$.Since $\vec{c} = \alpha\vec{a} + \beta\vec{b}$,$\vec{c} \cdot (\vec{a} + \vec{b}) = (\alpha\vec{a} + \beta\vec{b}) \cdot (\vec{a} + \vec{b}) = \alpha|\vec{a}|^2 + \beta|\vec{b}|^2 + (\alpha + \beta)(\vec{a} \cdot \vec{b})$.$|\vec{a}|^2 = 4+1+1 = 6$,$|\vec{b}|^2 = 1+4+1 = 6$,$\vec{a} \cdot \vec{b} = 2+2-1 = 3$.So,$6\alpha + 6\beta + 3(\alpha + \beta) = 9(\alpha + \beta)$.Thus,$\frac{9(\alpha + \beta)}{3\sqrt{2}} = 3\sqrt{2} \implies 9(\alpha + \beta) = 18 \implies \alpha + \beta = 2$.Now,$(\vec{c} - (\vec{a} 	imes \vec{b})) \cdot \vec{c} = |\vec{c}|^2 - (\vec{a} 	imes \vec{b}) \cdot \vec{c}$.Since $\vec{c} = \alpha\vec{a} + \beta\vec{b}$,$(\vec{a} 	imes \vec{b}) \cdot \vec{c} = 0$ because $\vec{c}$ is in the plane of $\vec{a}$ and $\vec{b}$.So,the expression is $|\vec{c}|^2 = |\alpha\vec{a} + \beta\vec{b}|^2 = \alpha^2|\vec{a}|^2 + \beta^2|\vec{b}|^2 + 2\alpha\beta(\vec{a} \cdot \vec{b}) = 6\alpha^2 + 6\beta^2 + 6\alpha\beta = 6(\alpha^2 + \beta^2 + \alpha\beta)$.Substitute $\beta = 2 - \alpha$: $6(\alpha^2 + (2-\alpha)^2 + \alpha(2-\alpha)) = 6(\alpha^2 + 4 - 4\alpha + \alpha^2 + 2\alpha - \alpha^2) = 6(\alpha^2 - 2\alpha + 4) = 6((\alpha - 1)^2 + 3)$.The minimum value is $6 	imes 3 = 18$.

Column-$I$	Column-$II$
$(A)$ In $R^2$,if the magnitude of the projection vector of the vector $\alpha \hat{i}+\beta \hat{j}$ on $\sqrt{3} \hat{i}+\hat{j}$ is $\sqrt{3}$ and if $\alpha=2+\sqrt{3} \beta$,then possible value$(s)$ of $\|\alpha\|$ is (are)	$(P)$ $1$
$(B)$ Let $a$ and $b$ be real numbers such that the function $f(x)=\begin{cases} -3ax^2-2, & x < 1 \\ bx+a^2, & x \geq 1 \end{cases}$ is differentiable for all $x \in R$. Then possible value$(s)$ of $a$ is (are)	$(Q)$ $2$
$(C)$ Let $\omega \neq 1$ be a complex cube root of unity. If $(3-3\omega+2\omega^2)^{4n+3} + (2+3\omega-3\omega^2)^{4n+3} + (-3+2\omega+3\omega^2)^{4n+3}=0$,then possible value$(s)$ of $n$ is (are)	$(R)$ $3$
$(D)$ Let the harmonic mean of two positive real numbers $a$ and $b$ be $4$. If $q$ is a positive real number such that $a, 5, q, b$ is an arithmetic progression,then the value$(s)$ of $\|q-a\|$ is (are)	$(S)$ $4$
	$(T)$ $5$

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