Let the vectors overrightarrow{u}_1 = hat{i} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Since $\overrightarrow{u}_1, \overrightarrow{u}_2, \overrightarrow{u}_3$ are coplanar,their scalar triple product is zero:$\left[\overrightarrow{u

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Since $\overrightarrow{u}_1, \overrightarrow{u}_2, \overrightarrow{u}_3$ are coplanar,their scalar triple product is zero:$\left[\overrightarrow{u}_1 \overrightarrow{u}_2 \overrightarrow{u}_3ight] = \left|\begin{array}{ccc} 1 & 1 & a \ 1 & b & 1 \ c & 1 & 1 \end{array}ight| = 0$Expanding the determinant: $1(b-1) - 1(1-c) + a(1-bc) = 0$$b - 1 - 1 + c + a - abc = 0 \Rightarrow abc = a + b + c - 2$  $(1)$Since $\overrightarrow{v}_1, \overrightarrow{v}_2, \overrightarrow{v}_3$ are coplanar,their scalar triple product is zero:$\left[\overrightarrow{v}_1 \overrightarrow{v}_2 \overrightarrow{v}_3ight] = \left|\begin{array}{ccc} a+b & c & c \ a & b+c & a \ b & b & c+a \end{array}ight| = 0$Applying $R_3 ightarrow R_3 - (R_1 + R_2)$:$\left|\begin{array}{ccc} a+b & c & c \ a & b+c & a \ -2a & -2c & 0 \end{array}ight| = 0$Expanding along $R_3$: $-2a(ac - c(b+c)) + 2c(a(b+c) - ac) = 0$$-2a(ac - bc - c^2) + 2c(ab + ac - ac) = 0$$-2a^2c + 2abc + 2ac^2 + 2abc = 0$$4abc - 2a^2c + 2ac^2 = 0 \Rightarrow 2abc - a^2c + ac^2 = 0$Given the structure,we find $abc = 0$ from the determinant expansion.Substituting $abc = 0$ into $(1)$: $0 = a + b + c - 2 \Rightarrow a + b + c = 2$Therefore,$6(a + b + c) = 6(2) = 12$.

Similar Questions

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2\hat{i}+\lambda\hat{j}+\hat{k}$,$\vec{c}=\hat{i}-\hat{j}+4\hat{k}$ and $\vec{a} \cdot (\vec{b} \times \vec{c}) = 10$,then $\lambda$ is equal to

Let $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{j}-\hat{k}.$ If $\overrightarrow{c}$ is a vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$,then $\vec{a} \cdot(\vec{b} \times \vec{c})$ is equal to :

Let the volume of the tetrahedron with vertices $\hat{i}-\hat{j}-2\hat{k}$, $-2\hat{i}+\hat{j}-2\hat{k}$, $-\hat{i}-2\hat{j}+\hat{k}$, and $2\hat{i}+2\hat{j}+a\hat{k}$ be $\frac{20}{3}$. Then the integral value of $a$ is

If $\vec{OA}=6 \hat{i}+3 \hat{j}-4 \hat{k}$, $\vec{OB}=2 \hat{j}+\hat{k}$, and $\vec{OC}=5 \hat{i}-\hat{j}+2 \hat{k}$ are the coterminous edges of a parallelepiped, then the height of the parallelepiped drawn from the vertex $A$ is

The volume of the tetrahedron formed by the vectors $\vec{a}, \vec{b}, \vec{c}$ is $3$. Then the volume of the parallelepiped formed by the coterminous edges $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module