$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
વચ્ચેનું ન્યૂનતમ અંતર $\frac{m}{\sqrt{n}}$ હોય, જ્યાં $\operatorname{gcd}(m, n)=1$, તો $m+n$ નું મૂલ્ય ........... છે.
$=\left|\frac{(0 \hat{i}+2 \hat{j}+2 \hat{k}) \cdot(-15 \hat{i}+7 \hat{j}+9 \hat{k})}{\sqrt{355}}\right|$
$=\frac{0+14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}}$
$\therefore \mathrm{m}+\mathrm{n}=32+355=387$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ One of the two boxes, box $I$ and box $II$, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $II$ is $\frac{1}{3}$, then the correct option$(s)$ with the possible values of $n_1, n_2, n_3$ and $n_4$ is(are)
$(A)$ $n_1=3, n_2=3, n_3=5, n_4=15$
$(B)$ $n_1=3, n_2=6, n_3=10, n_4=50$
$(C)$ $n_1=8, n_2=6, n_3=5, n_4=20$
$(D)$ $n_1=6, n_2=12, n_3=5, n_4=20$
$2.$ A ball is drawn at random from box $I$ and transferred to box $II$. If the probability of drawing a red ball from box $I$, after this transfer, is $\frac{1}{3}$, then the correct option$(s)$ with the possible values of $n_1$ and $n_2$ is(are)
$(A)$ $n_1=4, n_2=6$ $(B)$ $n_1=2, n_2=3$
$(C)$ $n_1=10, n_2=20$ $(D)$ $n_1=3, n_2=6$
Give the answer question $1$ and $2.$
$( S 1) P \left( A ^{\prime} \cup B \right)=\frac{5}{6}$
$( S 2) P \left( A ^{\prime} \cap B ^{\prime}\right)=\frac{1}{18}$. Then.
$x+y+3 z=0$
$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
માટે શૂન્યેતર ઉકેલ $(x, y, z)$ જ્યાં $k \in R$ હોય તો $x +\left(\frac{ y }{ z }\right)$ ની કિમત મેળવો
$ \leftarrow \,n\,\,times\, \to $