Let the vectors \(\mathbf{u}_1 = \hat{i} + \hat{j} + a\hat{k}, \mathbf{u}_2 = \hat{i} + b\hat{j} + \hat{k}\), and \(\mathbf{u}_3 = c\hat{i} + \hat{j} + \hat{k}\) be coplanar. If the vectors \(\mathbf{v}_1 = (a + b)\hat{i} + c\hat{j} + c\hat{k}, \mathbf{v}_2 = a\hat{i} + (b + c)\hat{j} + a\hat{k}, \mathbf{v}_3 = b\hat{i} + b\hat{j} + (c + a)\hat{k}\) are also coplanar, then \(6(a + b + c)\) is equal to:
If the points with position vectors \(a\hat{i} +10\hat{j} +13\hat{k}, 6\hat{i} +11\hat{k} +11\hat{k},\frac{9}{2}\hat{i}+B\hat{j}−8\hat{k}\) are collinear, then (19α-6β)2 is equal to
A molecule with the formula $ \text{A} \text{X}_2 \text{Y}_2 $ has all it's elements from p-block. Element A is rarest, monotomic, non-radioactive from its group and has the lowest ionization energy value among X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
A transition metal (M) among Mn, Cr, Co, and Fe has the highest standard electrode potential $ M^{n}/M^{n+1} $. It forms a metal complex of the type $[M \text{CN}]^{n+}$. The number of electrons present in the $ e $-orbital of the complex is ... ...
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.
Consider the following equilibrium, $$ \text{CO(g)} + \text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} $$ 0.1 mol of CO along with a catalyst is present in a 2 dm$^3$ flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH$_3$OH is formed. The $ K_p $ is ...... x $ 10^7 $ (nearest integer).
Given: $ R = 0.08 \, \text{dm}^3 \, \text{bar} \, \text{K}^{-1} \, \text{mol}^{-1} $
Assume only methanol is formed as the product and the system follows ideal gas behavior.
A physical quantity, represented both in magnitude and direction can be called a vector.
For the supplemental purposes of these vectors, there are two laws that are as follows;
It means that if we have any two vectors a and b, then for them
\(\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{b}+\overrightarrow{a}\)
It means that if we have any three vectors namely a, b and c.
\((\overrightarrow{a}+\overrightarrow{b})+\overrightarrow{c}=\overrightarrow{a}+(\overrightarrow{b}+\overrightarrow{c})\)
Read More: Addition of Vectors