Mathematics Test-5

Perpendicular Distance of a Point from a Plane

Let us consider a plane given by the Cartesian equation,

\(Ax+By+Cz=d\) and a point whose coordinate is,

\(\left(x_1, y_1, z_1\right)\).

Then the distance between the point and the plane is given by:

\(\left|\frac{A x_1+B y_1+C z_1-d}{\sqrt{A^2+B^2+C^2}}\right|\)

Quadratic equation with roots \(\alpha\) and \(\beta\) :

Quadratic equation with roots \(\alpha\) and \(\beta\) is given by:

\((x-\alpha)(x-\beta)=0\)

We know that the distance between the point and the plane is given by

\(\left|\frac{A x_1+B y_1+C z_1-d}{\sqrt{A^2+B^2+C^2}}\right|\)

So, the distance of the point \((2,3,-5)\) from the given plane

\(\alpha=\left|\frac{2 \times 1+2 \times 3+(-2) \times(-5)-9}{\sqrt{1^2+(-2)^2+(2)^2}}\right|=3\)

Similarly, the distance of the point \((3,1,1)\) from the given plane

\(\beta=\left|\frac{3 \times 1+2 \times 1+(-2) \times 1-9}{\sqrt{1^2+(-2)^2+(2)^2}}\right|=2\)

Therefore, quadratic equation with roots 3 & 2 will be

\((x-2)(x-3)=0\)

\(\Rightarrow x^2-3 x-2 x+6=0\)

\(\Rightarrow x^2-5 x+6=0\)

We know that the total number of symmetric relation in a set is given by \(2^{\frac{n(n+1)}{2}}\)where \(n\) is the number ofelements in the set.

So, let us put this formula and in place of \(n\) we will put 7 as there are a total of 7 elements in the given set.

\(\therefore 2^{\frac{n(n+1)}{2}}\)

\(=2^{\frac{7(7+1)}{2}}\)

\(=2^{\frac{7 \times 8}{2}}\)

\(=2^{7 \times 4}\)

\(=2^{28}\)

\(A=\left[\begin{array}{cc}x & -7 \\ 7 & y\end{array}\right]\)

\(\operatorname{Now} A^T=\left[\begin{array}{cc}x & 7 \\ -7 & y\end{array}\right]\)

Now for \(A\) to be skew symmetric matrix, \(A+A^T=0\)

\(\Rightarrow\left[\begin{array}{cc} x & -7 \\ 7 & y \end{array}\right]+\left[\begin{array}{cc} x & 7 \\ -7 & y \end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)

\(\Rightarrow\left[\begin{array}{cc}2 x & 0 \\ 0 & 2 y \end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)

Now equating the corresponding elements we get, \(( x , y )=(0,0)\)

Sum of Gp series of nth terms:

\(S_{n}=\frac{a \times\left(r^{n}-1\right)}{r-1}\)

Where,

a: First Term

r: common ratio

If \(\omega\) is a complex root of the unit. Then,

\(\omega^{3}=1\)

According to question

\(1+\omega+\omega^{2}+\ldots+\omega^{100}\)

We know that,

\(\omega^{3}=1\)

so, \(\left(\omega^{3}\right)^{33}=(1)^{33}\)

\(\omega^{99}=1\)

Now, Solving GP:

\(=\frac{1 \times\left(\omega^{101}-1\right)}{\omega-1}\)

\(=\frac{\omega^{99}\left(\omega^{2}-1\right)}{\omega-1}\)

\(=\frac{\omega^{99}(\omega-1)(\omega+1)}{\omega-1}\)

\(=\omega+1\)

Let the number of blue balls be \(x\).

Then total number of balls will be \(5+x\).

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

now,

the probability of drawing a blue ball is double that of a red ball, then the number of blue balls in a bag is:

Probability of drawing a blue ball \(= \frac{x}{5+x}\)

Probability of drawing a red ball \(= \frac{5}{5+x}\)

as given in question,

\(\Rightarrow\)\(\frac{x}{(5+x)}\) \(=2 \times (\frac{5}{5+x}) \)

\(\Rightarrow\) \( x =\) \( 2 \times5\)

\(\Rightarrow x=10\)

Concept:

Two matrices \(A\) and \(B\) are said to be equal if:

- They are of same order

- The corresponding elements of the matrices are same.

Calculation:

Comparing the corresponding elements:

\(M_{11}: a-b=-1 \Rightarrow b=a+1 \)

\(M_{21}: 2 a-b=0 \Rightarrow b=2 a \)

\(\Rightarrow 2 a=a+1 \Rightarrow a=1, b=2 \)

\(M_{12}: 2 a+c=5 \Rightarrow 2 \times 1+c=5 \Rightarrow c=3 \)

\(M_{22}: 3 c+d=13 \Rightarrow 3 \times 3+d=13 \Rightarrow d=4 \)

\(\therefore a=1, b=1, c=3, d=4\)

\(\tan ^2 \theta=\frac{8}{7}\)

\(\Rightarrow \tan \theta=\frac{\sqrt{8}}{\sqrt{7}}=\frac{2 \sqrt{2}}{\sqrt{7}}\)\(=\frac{\text { Perpendicular }}{\text { Base }}\)

By Pythagoras Theorem,

\((\text {Hypotenuse})^2=(\text {Base})^2+(\text { Perpendicular})^2 \)

\(=(\sqrt{7})^2+(2 \sqrt{2})^2=7+8=15 \)

\(\therefore \text { Hypotenuse}=\sqrt{15}\)

\(\therefore\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{2 \sqrt{2}}{\sqrt{15}} \)

\( \cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{\sqrt{7}}{\sqrt{15}}\)

\(=\frac{1-\sin ^2 \theta}{1-\cos ^2 \theta}=\frac{1-\left(\frac{2 \sqrt{2}}{\sqrt{15}}\right)^2}{1-\left(\frac{\sqrt{7}}{\sqrt{15}}\right)^2}\)

\(=\frac{1-\frac{8}{15}}{1-\frac{7}{15}}=\frac{\frac{15-8}{15}}{\frac{15-7}{15}}=\frac{\frac{7}{15}}{\frac{8}{15}}\)

\(=\frac{7}{15} \times \frac{15}{8}=\frac{7}{8}\)

Let angle between \(a\) and \(b\) is \(\theta\).

\(\vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cdot \cos \theta\) (Dot product of two vectors)

\(\vec{a} \cdot \vec{b}=\cos \theta\)

As \(\vec{ a }\) and \(\vec{ b }\) are unit vector so, \(|\vec{ a }|=|\vec{ b }|=1\).

\(\sqrt{3} \vec{a} - b\) is also unit vector i.e. \(|\sqrt{3} \vec{a}-\vec{b}|=1\)

Squaring both sides, we get,

\((\sqrt{3} \vec{a}-\vec{b})^{2}=1\)

\((\sqrt{3})^{2}|\vec{a}|^{2}+|\vec{b}|^{2}-2 \cdot \sqrt{3} \cdot|\vec{a} \cdot \vec{b}|=1 \)

\(\Rightarrow 3 \cdot 1+1-2 \cdot \sqrt{3} \cdot \cos \theta=1 \)

[Since, \(\vec{a} \cdot \vec{b}=\cos \theta \)]

\(\Rightarrow 4-2 \sqrt{3} \cdot \cos \theta=1 \)

\(\Rightarrow 2 \sqrt{3} \cdot \cos \theta=3 \)

\(\Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \)

\(\Rightarrow \theta=\frac{\pi}{6}\)

Therefore, the angle between the two unit vectors is \(\frac{\pi}{6}\).

Given,

\(\sum_{r=0}^{50}\left({ }^{50} C_{r}\right)\left(3^{50-r}\right)(x-2)^{r}\)

\(⇒ (3+(x-2))^{50}\)

\(⇒ (x+1)^{50}\)

Coefficient of \(x^{48}\) in \((1+x)^{50}\) is

\({ }^{50} C_{48}={ }^{50} C_{2}\)