Answer

Verified

431.1k+ views

Hint: The three vectors given to us are coplanar means the determinant consists of the coefficients of the vector is zero. Another condition is given to us that is the absolute value of the vector c is given. Using these two conditions we can find out the required values.

Complete step-by-step answer:

A vector is a line segment that has a length and definite direction.

The vectors which are parallel to the same plane, or lie on the same plane are called coplanar vectors.

If any three vectors are linearly dependent then they are coplanar.

It is given in the question that the three vectors $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=4\hat{i}+3\hat{j}+4\hat{k},\overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$ are coplanar.

That means they are linearly dependent vectors.

Therefore the determinant of the coefficients of these three vectors will be equal to zero.

$\begin{align}

& \left| \begin{matrix}

1 & 1 & 1 \\

4 & 3 & 4 \\

1 & \alpha & \beta \\

\end{matrix} \right|=0 \\

& \Rightarrow 1\times \left( 3\beta -4\alpha \right)-1\times \left( 4\beta -4 \right)+1\times \left( 4\alpha -3 \right)=0 \\

& \Rightarrow 3\beta -4\alpha -4\beta +4+4\alpha -3=0 \\

& \Rightarrow -\beta +1=0 \\

& \Rightarrow \beta =1 \\

\end{align}$

So by using the first condition we got the value of $\beta $.

Now we have to find out the value of $\alpha $.

We have another condition that is given in the question.

Which is the absolute value of the third vector is given to us.

The absolute value or the modulus value is the measure of the magnitude of a vector. Geometrically, the absolute value represents the displacement from the origin and it is always nonnegative.

We have,

$\left| c \right|=\sqrt{3}$

By squaring both the sides we have,

$\begin{align}

& \Rightarrow {{\left| c \right|}^{2}}={{\left( \sqrt{3} \right)}^{2}} \\

& \Rightarrow {{\left( \sqrt{{{1}^{2}}+{{\alpha }^{2}}+{{\beta }^{2}}} \right)}^{2}}=3 \\

& \Rightarrow 1+{{\alpha }^{2}}+{{\beta }^{2}}=3 \\

& \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=3-1=2 \\

\end{align}$

Substitute the value of $\beta $,

$\begin{align}

& \Rightarrow {{\alpha }^{2}}+{{\left( 1 \right)}^{2}}=2 \\

& \Rightarrow {{\alpha }^{2}}=2-1=1 \\

& \Rightarrow \alpha =\pm 1 \\

\end{align}$

Therefore the values are $\alpha =\pm 1,\beta =1$.

Hence, option (c) is correct.

Note: Alternatively we can use the fact that three vectors are said to be coplanar if their scalar triple product is zero.

Therefore, $\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=0$

We can get the value of $\beta $ from the above condition.

Complete step-by-step answer:

A vector is a line segment that has a length and definite direction.

The vectors which are parallel to the same plane, or lie on the same plane are called coplanar vectors.

If any three vectors are linearly dependent then they are coplanar.

It is given in the question that the three vectors $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=4\hat{i}+3\hat{j}+4\hat{k},\overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$ are coplanar.

That means they are linearly dependent vectors.

Therefore the determinant of the coefficients of these three vectors will be equal to zero.

$\begin{align}

& \left| \begin{matrix}

1 & 1 & 1 \\

4 & 3 & 4 \\

1 & \alpha & \beta \\

\end{matrix} \right|=0 \\

& \Rightarrow 1\times \left( 3\beta -4\alpha \right)-1\times \left( 4\beta -4 \right)+1\times \left( 4\alpha -3 \right)=0 \\

& \Rightarrow 3\beta -4\alpha -4\beta +4+4\alpha -3=0 \\

& \Rightarrow -\beta +1=0 \\

& \Rightarrow \beta =1 \\

\end{align}$

So by using the first condition we got the value of $\beta $.

Now we have to find out the value of $\alpha $.

We have another condition that is given in the question.

Which is the absolute value of the third vector is given to us.

The absolute value or the modulus value is the measure of the magnitude of a vector. Geometrically, the absolute value represents the displacement from the origin and it is always nonnegative.

We have,

$\left| c \right|=\sqrt{3}$

By squaring both the sides we have,

$\begin{align}

& \Rightarrow {{\left| c \right|}^{2}}={{\left( \sqrt{3} \right)}^{2}} \\

& \Rightarrow {{\left( \sqrt{{{1}^{2}}+{{\alpha }^{2}}+{{\beta }^{2}}} \right)}^{2}}=3 \\

& \Rightarrow 1+{{\alpha }^{2}}+{{\beta }^{2}}=3 \\

& \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=3-1=2 \\

\end{align}$

Substitute the value of $\beta $,

$\begin{align}

& \Rightarrow {{\alpha }^{2}}+{{\left( 1 \right)}^{2}}=2 \\

& \Rightarrow {{\alpha }^{2}}=2-1=1 \\

& \Rightarrow \alpha =\pm 1 \\

\end{align}$

Therefore the values are $\alpha =\pm 1,\beta =1$.

Hence, option (c) is correct.

Note: Alternatively we can use the fact that three vectors are said to be coplanar if their scalar triple product is zero.

Therefore, $\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=0$

We can get the value of $\beta $ from the above condition.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What organs are located on the left side of your body class 11 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is pollution? How many types of pollution? Define it