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# If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=4\hat{i}+3\hat{j}+4\hat{k},\overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$ are coplanar and $\left| c \right|=\sqrt{3}$, then:(a).$\alpha =\sqrt{2},\beta =1$ (b).$\alpha =1,\beta =\pm 1$ (c).$\alpha =\pm 1,\beta =1$ (d).$\alpha =\pm 1,\beta =-1$ (e).$\alpha =-1,\beta =\pm 1$

Last updated date: 23rd Mar 2023
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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.

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.