Find the value of x such …

Ans.

Points are A(3,2,1), B(4,x,5), C(4,2,-2) And D(6,5,-1).

Position Vector of A = {tex}3\hat i + 2\hat j + \hat k{/tex}

Position Vector of B = {tex}4\hat i + x\hat j + 5\hat k{/tex}

Position Vector of C = {tex}4\hat i + 2\hat j -2\hat k{/tex}

Position Vector of D = {tex}6\hat i + 5\hat j -\hat k{/tex}

{tex}=> \overrightarrow {AB} = \hat i + (2-x)\hat j + 4\hat k{/tex}

{tex}=> \overrightarrow {AC} = \hat i + 0\hat j -3\hat k{/tex} 

{tex}=> \overrightarrow {AD} = 3\hat i + 3\hat j -2\hat k{/tex}

As points are coplaner, then triple scaler product is zero,

So,

{tex}\begin{bmatrix} 1& (x-2) & 4 \\[0.3em] 1 & 0 & -3 \\[0.3em] 3 & 3 & -2 \end{bmatrix} = 0{/tex}

Expanding along R₂

=> -1[-2(x-2) -12] -(-3)[3-3(x-2)] = 0

=> 2x -4+12 +9-9x +18 = 0

=> -7x +35 = 0

=> 7x = 35

=> x = 5