Question

If $a,b,c$ are distinct positive numbers and the vectors $a\hat i+a\hat j+c\hat k,\ \hat i+\hat k$ and $c\hat i+c\hat j+b\hat k$ lie in a plane, then

• A. $c$ is A.M. of $a$ and $b$
• B. $c^2=0$
• C. $c$ is H.M. of $a$ and $b$
• D. $c$ is G.M. of $a$ and $b$

Hint:

Coplanarity of vectors is tested using the scalar triple product.

Answer 1

The Correct Option is D

Step 1: Writing the given vectors.
\[ \vec v_1=\langle a,a,c\rangle,\quad \vec v_2=\langle 1,0,1\rangle,\quad \vec v_3=\langle c,c,b\rangle \]
Step 2: Using coplanarity condition.
Three vectors are coplanar if \[ \begin{vmatrix} a & a & c
1 & 0 & 1
c & c & b \end{vmatrix}=0 \]
Step 3: Evaluating the determinant.
\[ a(0\cdot b-1\cdot c)-a(1\cdot b-1\cdot c)+c(1\cdot c-0\cdot c)=0 \] \[ -ac-a(b-c)+c^2=0 \] \[ c^2=ab \]
Step 4: Conclusion.
Since $c^2=ab$, \[ c=\sqrt{ab} \] Thus, $c$ is the geometric mean of $a$ and $b$.