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# A uni-modular tangent vector on the curve $x={{t}^{2}}+2,y=4t-5,z=2{{t}^{2}}-6t$ at $t=2$ is \begin{align} & (\text{A) }\dfrac{1}{3}\left( 2\overset{\scriptscriptstyle\rightharpoonup}{i}+2\overset{\scriptscriptstyle\rightharpoonup}{j}+\overset{\scriptscriptstyle\rightharpoonup}{k} \right) \\ & (B\text{) }\dfrac{1}{3}\left( \overset{\scriptscriptstyle\rightharpoonup}{i}-\overset{\scriptscriptstyle\rightharpoonup}{j}-\overset{\scriptscriptstyle\rightharpoonup}{k} \right) \\ & (C\text{) }\dfrac{1}{3}\left( 2\overset{\scriptscriptstyle\rightharpoonup}{i}+\overset{\scriptscriptstyle\rightharpoonup}{j}+\overset{\scriptscriptstyle\rightharpoonup}{k} \right) \\ & (D\text{) }\dfrac{2}{3}\left( \overset{\scriptscriptstyle\rightharpoonup}{i}+\overset{\scriptscriptstyle\rightharpoonup}{j}+\overset{\scriptscriptstyle\rightharpoonup}{k} \right) \\ \end{align}

Last updated date: 23rd May 2024
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Hint: We know that if $f(x,y,z,t)=0$ represents a curve then the tangent vector of $f(x,y,z,t)=0$ is represented by $\dfrac{dx}{dt}\overset{\scriptscriptstyle\rightharpoonup}{i}+\dfrac{dy}{dt}\overset{\scriptscriptstyle\rightharpoonup}{j}+\dfrac{dz}{dt}\overset{\scriptscriptstyle\rightharpoonup}{k}$. From the question, we were given the equation of curve is $x={{t}^{2}}+2,y=4t-5,z=2{{t}^{2}}-6t$. Now we have to calculate $\dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}$. From the values of $\dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}$ we can get the vector $\dfrac{dx}{dt}\overset{\scriptscriptstyle\rightharpoonup}{i}+\dfrac{dy}{dt}\overset{\scriptscriptstyle\rightharpoonup}{j}+\dfrac{dz}{dt}\overset{\scriptscriptstyle\rightharpoonup}{k}$. Now we have to substitute $t=2$. This will give us the tangent vector of $f(x,y,z,t)=0$ at $t=2$. We know that the unit vector of $a\overset{\scriptscriptstyle\rightharpoonup}{i}+b\overset{\scriptscriptstyle\rightharpoonup}{j}+c\overset{\scriptscriptstyle\rightharpoonup}{k}$ is $\left( \dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{i}+\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{j}+\left( \dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{k}$. Now by using this concept, we can find the uni-modular tangent vector of curve $x={{t}^{2}}+2,y=4t-5,z=2{{t}^{2}}-6t$ at $t=2$.

Complete step-by-step solution:
Before solving the question, we should know that if $f(x,y,z,t)=0$ represents a curve then the tangent vector of $f(x,y,z,t)=0$ is represented by $\dfrac{dx}{dt}\overset{\scriptscriptstyle\rightharpoonup}{i}+\dfrac{dy}{dt}\overset{\scriptscriptstyle\rightharpoonup}{j}+\dfrac{dz}{dt}\overset{\scriptscriptstyle\rightharpoonup}{k}$.
From the question, we were given the equation of curve is $x={{t}^{2}}+2,y=4t-5,z=2{{t}^{2}}-6t$.
Let us consider
\begin{align} & x={{t}^{2}}+2......(1) \\ & y=4t-5.......(2) \\ & z=2{{t}^{2}}-6t......(3) \\ \end{align}
Now let us differentiate equation (1) on both sides, then we get
\begin{align} & \Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{2}}+2 \right) \\ & \Rightarrow \dfrac{dx}{dt}=2t.....(4) \\ \end{align}
Now let us differentiate equation (2) on both sides, then we get
\begin{align} & \Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( 4t-5 \right) \\ & \Rightarrow \dfrac{dy}{dt}=4.....(5) \\ \end{align}
Now let us differentiate equation (3) on both sides, then we get
\begin{align} & \Rightarrow \dfrac{dz}{dt}=\dfrac{d}{dt}\left( 2{{t}^{2}}-6t \right) \\ & \Rightarrow \dfrac{dz}{dt}=4t-6.....(6) \\ \end{align}
From equation (4), equation (5) and equation (6), we can get the tangent vector of the curve $x={{t}^{2}}+2,y=4t-5,z=2{{t}^{2}}-6t$.
Now we have to find the equation of the tangent vector.
$\Rightarrow \dfrac{dx}{dt}\overset{\scriptscriptstyle\rightharpoonup}{i}+\dfrac{dy}{dt}\overset{\scriptscriptstyle\rightharpoonup}{j}+\dfrac{dz}{dt}\overset{\scriptscriptstyle\rightharpoonup}{k}=\left( 2t \right)\overset{\scriptscriptstyle\rightharpoonup}{i}+4\overset{\scriptscriptstyle\rightharpoonup}{j}+\left( 4t-6 \right)\overset{\scriptscriptstyle\rightharpoonup}{k}.....(7)$
Now we have to find the equation of the tangent at $t=2$.
So, now we have to substitute $t=2$ in equation (7), then we get
\begin{align} & \Rightarrow \dfrac{dx}{dt}\overset{\scriptscriptstyle\rightharpoonup}{i}+\dfrac{dy}{dt}\overset{\scriptscriptstyle\rightharpoonup}{j}+\dfrac{dz}{dt}\overset{\scriptscriptstyle\rightharpoonup}{k}=\left( 2(2) \right)\overset{\scriptscriptstyle\rightharpoonup}{i}+4\overset{\scriptscriptstyle\rightharpoonup}{j}+\left( 4(2)-6 \right)\overset{\scriptscriptstyle\rightharpoonup}{k} \\ & \Rightarrow \dfrac{dx}{dt}\overset{\scriptscriptstyle\rightharpoonup}{i}+\dfrac{dy}{dt}\overset{\scriptscriptstyle\rightharpoonup}{j}+\dfrac{dz}{dt}\overset{\scriptscriptstyle\rightharpoonup}{k}=4\overset{\scriptscriptstyle\rightharpoonup}{i}+4\overset{\scriptscriptstyle\rightharpoonup}{j}+2\overset{\scriptscriptstyle\rightharpoonup}{k}.....(8) \\ \end{align}
Now we have to find the unimodular vector of equation (8).
We know that the unit vector of $a\overset{\scriptscriptstyle\rightharpoonup}{i}+b\overset{\scriptscriptstyle\rightharpoonup}{j}+c\overset{\scriptscriptstyle\rightharpoonup}{k}$ is $\left( \dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{i}+\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{j}+\left( \dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{k}$.
Now let us compare $4\overset{\scriptscriptstyle\rightharpoonup}{i}+4\overset{\scriptscriptstyle\rightharpoonup}{j}+2\overset{\scriptscriptstyle\rightharpoonup}{k}$ with $a\overset{\scriptscriptstyle\rightharpoonup}{i}+b\overset{\scriptscriptstyle\rightharpoonup}{j}+c\overset{\scriptscriptstyle\rightharpoonup}{k}$. Then we get
\begin{align} & a=4......(9) \\ & b=4......(10) \\ & c=2.......(11) \\ \end{align}
Now from equation (9), equation (10) and equation (11) we have to calculate the value of $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.
\begin{align} & \Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=\sqrt{{{4}^{2}}+{{4}^{2}}+{{2}^{2}}} \\ & \Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=\sqrt{36} \\ & \Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=6.....(12) \\ \end{align}
Now we have to find the unit tangent vector.
From equation (9), equation (10), equation (11) and equation (12), we get
\begin{align} & \Rightarrow \left( \dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{i}+\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{j}+\left( \dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{k}=\left( \dfrac{4}{6} \right)\overset{\scriptscriptstyle\rightharpoonup}{i}+\left( \dfrac{4}{6} \right)\overset{\scriptscriptstyle\rightharpoonup}{j}+\left( \dfrac{2}{6} \right)\overset{\scriptscriptstyle\rightharpoonup}{k} \\ & \Rightarrow \left( \dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{i}+\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{j}+\left( \dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{k}=\dfrac{4\overset{\scriptscriptstyle\rightharpoonup}{i}+4\overset{\scriptscriptstyle\rightharpoonup}{j}+2\overset{\scriptscriptstyle\rightharpoonup}{k}}{6} \\ & \Rightarrow \left( \dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{i}+\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{j}+\left( \dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right)\overset{\scriptscriptstyle\rightharpoonup}{k}=\dfrac{2\overset{\scriptscriptstyle\rightharpoonup}{i}+2\overset{\scriptscriptstyle\rightharpoonup}{j}+\overset{\scriptscriptstyle\rightharpoonup}{k}}{3}......(13) \\ \end{align}
From equation we can say that uni-modular tangent vector on the curve $x={{t}^{2}}+2,y=4t-5,z=2{{t}^{2}}-6t$ at $t=2$ is $\dfrac{2\overset{\scriptscriptstyle\rightharpoonup}{i}+2\overset{\scriptscriptstyle\rightharpoonup}{j}+\overset{\scriptscriptstyle\rightharpoonup}{k}}{3}$.
Hence, option A is correct.

Note: Students may do the solution up to equation (8) and they may conclude that $4\overset{\scriptscriptstyle\rightharpoonup}{i}+4\overset{\scriptscriptstyle\rightharpoonup}{j}+2\overset{\scriptscriptstyle\rightharpoonup}{k}$ is the tangent vector. They may think that this is the correct answer but this will give us the wrong answer. The reason is that in the question it was given that to find the uni-modular vector. We know that $4\overset{\scriptscriptstyle\rightharpoonup}{i}+4\overset{\scriptscriptstyle\rightharpoonup}{j}+2\overset{\scriptscriptstyle\rightharpoonup}{k}$ will not represents a uni -modular vector. So, students should follow the question properly and then proceed through the solution.