If the points with position vectors hat{i}-2 | vedclass

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

Question

(B) Let $A, B, C$, and $D$ be the points with position vectors $\vec{A} = \hat{i}-2 \hat{j}+3 \hat{k}$, $\vec{B} = 2 \hat{i}+3 \hat{j}-4 \hat{k}$, $\v

Vedclass · Accepted Answer

$\frac{42}{19}$

Vedclass · Answer

(B) Let $A, B, C$, and $D$ be the points with position vectors $\vec{A} = \hat{i}-2 \hat{j}+3 \hat{k}$, $\vec{B} = 2 \hat{i}+3 \hat{j}-4 \hat{k}$, $\vec{C} = -3 \hat{i}+\hat{j}-5 \hat{k}$, and $\vec{D} = a \hat{i}-2 \hat{j}+4 \hat{k}$.The vectors formed are:$\overrightarrow{AB} = \vec{B} - \vec{A} = (2-1)\hat{i} + (3-(-2))\hat{j} + (-4-3)\hat{k} = \hat{i} + 5\hat{j} - 7\hat{k}$$\overrightarrow{AC} = \vec{C} - \vec{A} = (-3-1)\hat{i} + (1-(-2))\hat{j} + (-5-3)\hat{k} = -4\hat{i} + 3\hat{j} - 8\hat{k}$$\overrightarrow{AD} = \vec{D} - \vec{A} = (a-1)\hat{i} + (-2-(-2))\hat{j} + (4-3)\hat{k} = (a-1)\hat{i} + 0\hat{j} + \hat{k}$Since the points are coplanar, the scalar triple product of these vectors must be zero:$[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}] = 0 \Rightarrow \begin{vmatrix} 1 & 5 & -7 \ -4 & 3 & -8 \ a-1 & 0 & 1 \end{vmatrix} = 0$Expanding along the third row:$(a-1) \begin{vmatrix} 5 & -7 \ 3 & -8 \end{vmatrix} - 0 \begin{vmatrix} 1 & -7 \ -4 & -8 \end{vmatrix} + 1 \begin{vmatrix} 1 & 5 \ -4 & 3 \end{vmatrix} = 0$$(a-1)(-40 - (-21)) + 1(3 - (-20)) = 0$$(a-1)(-19) + 23 = 0$$-19a + 19 + 23 = 0$$-19a + 42 = 0$$19a = 42 \Rightarrow a = \frac{42}{19}$

If the points with position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$, $2 \hat{i}+3 \hat{j}-4 \hat{k}$, $-3 \hat{i}+\hat{j}-5 \hat{k}$, and $a \hat{i}-2 \hat{j}+4 \hat{k}$ are coplanar, then $a=$

Similar Questions

If $3 \hat{i}-2 \hat{j}-\hat{k}$,$2 \hat{i}+3 \hat{j}-4 \hat{k}$,$-\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are respectively the position vectors of four coplanar points $P, Q, R$ and $S$,then $\lambda=$

If $\bar{a}, \bar{b}$ and $\bar{c}$ are any three non-zero vectors,then $(\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]=$

Volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is $12$ cubic units. The volume of a tetrahedron with coterminous edges $\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{a} + \vec{b} - \vec{c}$ will be ............. cubic units.

If the vectors $i+3j-2k$,$2i-j+4k$ and $3i+2j+xk$ are coplanar,then the value of $x$ is:

If $[\bar{a} \quad \bar{b} \quad \bar{c}]=4$,then the volume of the parallelepiped with coterminous edges $\bar{a}+2 \bar{b}$,$\bar{b}+2 \bar{c}$,and $\bar{c}+2 \bar{a}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module