Consider the four points A(1, -2, -1), B(4, 0 | vedclass

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

Question

(D) Given points are $A(1, -2, -1)$, $B(4, 0, -3)$, $C(1, 2, -1)$, and $D(2, -4, -5)$.The vectors are:$\vec{b} = \vec{AB} = (4-1)\hat{i} + (0-(-2))\ha

Vedclass · Accepted Answer

$-20$

Vedclass · Answer

(D) Given points are $A(1, -2, -1)$, $B(4, 0, -3)$, $C(1, 2, -1)$, and $D(2, -4, -5)$.The vectors are:$\vec{b} = \vec{AB} = (4-1)\hat{i} + (0-(-2))\hat{j} + (-3-(-1))\hat{k} = 3\hat{i} + 2\hat{j} - 2\hat{k}$$\vec{c} = \vec{AC} = (1-1)\hat{i} + (2-(-2))\hat{j} + (-1-(-1))\hat{k} = 0\hat{i} + 4\hat{j} + 0\hat{k}$$\vec{d} = \vec{AD} = (2-1)\hat{i} + (-4-(-2))\hat{j} + (-5-(-1))\hat{k} = 1\hat{i} - 2\hat{j} - 4\hat{k}$The scalar triple product $[\vec{b}, \vec{c}, \vec{d}] = \begin{vmatrix} 3 & 2 & -2 \ 0 & 4 & 0 \ 1 & -2 & -4 \end{vmatrix} = 3(-16) - 2(0) - 2(-4) = -48 + 8 = -40$.We know that $[\vec{b} 	imes \vec{c}, \vec{c} 	imes \vec{d}, \vec{d} 	imes \vec{b}] = [\vec{b}, \vec{c}, \vec{d}]^2 = (-40)^2 = 1600$.Also, $[\vec{b}+\vec{c}, \vec{c}+\vec{d}, \vec{d}+\vec{b}] = 2[\vec{b}, \vec{c}, \vec{d}] = 2(-40) = -80$.Therefore, the required value is $\frac{1600}{-80} = -20$.

Similar Questions

If $a = 3i - 2j + 2k$,$b = 6i + 4j - 2k$ and $c = 3i - 2j - 4k$,then $a \cdot (b \times c)$ is

The vector $(\hat{i} \times \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} \times \vec{a} \cdot \vec{b})\hat{j} + (\hat{k} \times \vec{a} \cdot \vec{b})\hat{k}$ is equal to

If $a, b, c$ are three coplanar vectors,then $[a + b, b + c, c + a] = $

The volume of a tetrahedron whose vertices are $4 \hat{i}+5 \hat{j}+\hat{k}$, $-\hat{j}+\hat{k}$, $3 \hat{i}+9 \hat{j}+4 \hat{k}$ and $-2 \hat{i}+4 \hat{j}+4 \hat{k}$ is (in cubic units)

For any non-zero vectors $a, b, c$,$a \cdot[(b+c) \times(a+b+c)] = \ldots .$

Vedclass Test Series

Exam Paper Generator

Online Exam Module