Jo Q13. Find the foot of the perpendicular from the point (1,2,0) upon the plane x-3y+27=9 and find the distance of the point (1,2,0) from the given point.
Answers
Answer:
Solution−
Given determinant is
\begin{gathered}\rm \: 2\left|\begin{array}{ll}\tt\sin (A+B) &\tt \cos (A+B) \\ \tt\cos (A-B) & \tt\sin (A-B)\end{array}\right|+\sqrt{3}=0 \\ \end{gathered}
2
∣
∣
∣
∣
∣
sin(A+B)
cos(A−B)
cos(A+B)
sin(A−B)
∣
∣
∣
∣
∣
+
3
=0
can be rewritten as
\begin{gathered}\rm \: 2\left|\begin{array}{ll}\tt\sin (A+B) &\tt \cos (A+B) \\ \tt\cos (A-B) & \tt\sin (A-B)\end{array}\right|= - \sqrt{3} \\ \end{gathered}
2
∣
∣
∣
∣
∣
sin(A+B)
cos(A−B)
cos(A+B)
sin(A−B)
∣
∣
∣
∣
∣
=−
3
\begin{gathered}\rm \: \left|\begin{array}{ll}\tt\sin (A+B) &\tt \cos (A+B) \\ \tt\cos (A-B) & \tt\sin (A-B)\end{array}\right|= -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}
∣
∣
∣
∣
∣
sin(A+B)
cos(A−B)
cos(A+B)
sin(A−B)
∣
∣
∣
∣
∣
=−
2
3
\begin{gathered}\rm \: sin(A + B)sin(A - B) - cos(A + B)cos(A - B)= -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}
sin(A+B)sin(A−B)−cos(A+B)cos(A−B)=−
2
3
\begin{gathered}\rm \:({sin}^{2}A- {sin}^{2}B)-( {cos}^{2}A-{sin}^{2}B) = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}
(sin
2
A−sin
2
B)−(cos
2
A−sin
2
B)=−
2
3
\begin{gathered}\rm \:{sin}^{2}A- {sin}^{2}B-{cos}^{2}A + {sin}^{2}B = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}
sin
2
A−sin
2
B−cos
2
A+sin
2
B=−
2
3
\begin{gathered}\rm \:{sin}^{2}A-{cos}^{2}A = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}
sin
2
A−cos
2
A=−
2
3
\begin{gathered}\rm \: - ( - {sin}^{2}A + {cos}^{2}A) = -\dfrac{ \sqrt{3} }{2} \\ \end{gathered}
−(−sin
2
A+cos
2
A)=−
2
3
\begin{gathered}\rm \: {cos}^{2}A - {sin}^{2}A = \dfrac{ \sqrt{3} }{2} \\ \end{gathered}
cos
2
A−sin
2
A=
2
3
\begin{gathered}\rm \: {cos}2A = \dfrac{ \sqrt{3} }{2} \\ \end{gathered}
cos2A=
2
3
\begin{gathered}\rm \: {cos}2A = cos\bigg(\dfrac{\pi}{6} \bigg) \\ \end{gathered}
cos2A=cos(
6
π
)
We know,
\begin{gathered}\red{\boxed{ \rm{ \:cosx = cosy \: \rm\implies \:x = 2n\pi \pm \: y\: \: \: \forall \: \: \: n \in \: Z}}}\\\end{gathered}
cosx=cosy⟹x=2nπ±y∀n∈Z
So, using this result, we get
\begin{gathered}\rm \: 2x = 2n\pi \pm \: \dfrac{\pi}{6} \: \: \: \forall \: n \in \: Z\\\end{gathered}
2x=2nπ±
6
π
∀n∈Z
\begin{gathered}\rm\implies \:\rm \: x = n\pi \pm \: \dfrac{\pi}{12} \: \: \: \forall \: n \in \: Z\\\end{gathered}
⟹x=nπ±
12
π
∀n∈Z
\rule{190pt}{2pt}
Formulae Used :-
\begin{gathered}\boxed{ \rm{ \:sin(A + B) \: sin(A - B) = {sin}^{2}A - {sin}^{2}B \: }} \\ \end{gathered}
sin(A+B)sin(A−B)=sin
2
A−sin
2
B
\begin{gathered}\boxed{ \rm{ \:cos(A + B) \: cos(A - B) = {cos}^{2}A - {sin}^{2}B \: }} \\ \end{gathered}
cos(A+B)cos(A−B)=cos
2
A−sin
2
B
\begin{gathered}\boxed{ \rm{ \:cos \frac{\pi}{6} = \frac{ \sqrt{3} }{2} \: }} \\ \end{gathered}
cos
6
π
=
2
3
\rule{190pt}{2pt}
Additional Information :-
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}
T−eq
sinx=0
cosx=0
tanx=0
sinx=siny
cosx=cosy
tanx=tany
Solution
x=nπ∀n∈Z
x=(2n+1)
2
π
∀n∈Z
x=nπ∀n∈Z
x=nπ+(−1)
n
y∀n∈Z
x=2nπ±y∀n∈Z
x=nπ+y∀n∈Z