Question

The vectors $\vec{a} = x \hat{i} + (x +1) \hat{j} + ( x +2 ) \hat{k}$ $\vec{b} = (x + 3) \hat{i} + (x +4) \hat{j} + ( x + 5 ) \hat{k}$ , and $\vec{c} = (x + 6) \hat{i} + (x + 7 ) \hat{j} + ( x + 8) \hat{k}$ are co-planar for

• A. an values of x
• B. x > 0
• C. x < 0
• D. none of these

Answer 1

Answer: (A)

an values of x

Answer 2

Vectors $\vec{a} = x \hat{i} + (x +1) \hat{j} + ( x +2 ) \hat{k}$
$\vec{b} = (x + 3) \hat{i} + (x +4) \hat{j} + ( x + 5 ) \hat{k}$ , and $\vec{c} = (x + 6) \hat{i} + (x + 7 ) \hat{j} + ( x + 8) \hat{k}$ are coplanar
$\therefore \:\:\: \left[\vec{a} .\vec{b} .\vec{c} \right] = 0$
$i.e. \vec{a} \left(\vec{b} \times \vec{c} \right) = 0$
$b \times c = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ x+3&x+4&x+5\\\ x+6&x+7&x+8\end{vmatrix}$
$= \hat{i} \left[x^{2} +12x +32 -x^{2} -12x -35\right]$
$-\hat{j}\left[x^{2} +11x+24 -x^{2}-11x-30\right]$
$+\hat{k}\left[x^{2} +10x+21 -x^{2} -10x -24\right]$
$=-3\hat{i } + 6\hat{j} -3\hat{k}$
So, $\vec{a}.\left(\vec{b} \times \vec{ c} \right)=-3x +6x +6 -3x -6 = 0$
Thus $\left[\vec{a} .\vec{b} .\vec{c} \right]$ = 0 for all values of x

The term "vectors" or "vector quantities" refers to quantities that have both magnitude and direction. The items known as vectors are those that may be discovered in cumulative form in vector spaces using two different types of operations. These operations in the vector space include multiplying the vector by a scalar quantity and adding two vectors together. The result still exists in the vector space even if these operations might change the vector's ratios and order. It is frequently identified by symbols like U, V, and W.

A directed line is one that has an arrowhead. The size and direction of a section of the directed line are both present. A vector is the name for this section of the directed line. It is shown as a or, more frequently, as AB. A and B are the line's beginning and ending points, respectively, in this segment of the line AB.