The position vectors of the vertices A,...

MathematicsEquation of a Line in SpaceJEE Advanced 1996Difficult

The position vectors of the vertices $A, B$ and $C$ of a tetrahedron $A B C D$ are $\hat{i} + \hat{j} + \hat{k}, \hat{i}$ and $3 \hat{i}$ , respectively. The altitude from vertex $D$ to the opposite face $A B C$ meets the median line through $A$ of the triangle $A B C$ at a point $E$ . If the length of the side $A D$ is $4$ and the volume of the tetrahedron is $\frac{2 2}{3}$ , find the position vector of the point $E$ for all its possible positions.

Answer:(-1, 3, 3) or (3, -1, -1)

Visualized Solution (Hindi)

Visualize the Tetrahedron .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A B C D

Vertices: $A (1, 1, 1)$ , $B (1, 0, 0)$ , $C (3, 0, 0)$
Face $A B C$ forms the base of the tetrahedron.
Goal: Find position vector of point $E$ on the median from $A$ .

Find the Median .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A M and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Midpoint $M$ of $B C$ : $M = (\frac{1 + 3}{2}, 0, 0) = (2, 0, 0)$
Direction of median $A M$ : $v = M - A = (1, - 1, - 1)$
Equation of line $A M$ : $r = (1, 1, 1) + t (1, - 1, - 1)$
Point $E$ on $A M$ : $E = (1 + t, 1 - t, 1 - t)$

Calculate Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ A B C

$A B = (0, - 1, - 1)$ , $A C = (2, - 1, - 1)$
$A B \times A C = \hat{i} 02 \hat{j} - 1 - 1 \hat{k} - 1 - 1 = (0, - 2, 2)$
Area of $△ A B C = \frac{1}{2} ∣ A B \times A C ∣ = \frac{1}{2} 0^{2} + (- 2)^{2} + 2^{2} = 2$

Find Altitude Height .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Volume $V = \frac{1}{3} \times Area (A B C) \times h$
$\frac{2 2}{3} = \frac{1}{3} \times 2 \times h$
Solving for $h$ : $h = 2$

Define Coordinates of Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Normal vector $n = (0, - 1, 1)$ , Unit normal $\overset{n}{^} = \frac{( 0 , - 1 , 1 )}{2}$
$D = E \pm h \overset{n}{^} = (1 + t, 1 - t, 1 - t) \pm 2 \frac{( 0 , - 1 , 1 )}{2}$
$D = (1 + t, 1 - t \mp 2, 1 - t \pm 2)$

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A D = 4

Given $A D = 4 ⟹ A D^{2} = 16$
$A D^{2} = (t)^{2} + (- t \mp 2)^{2} + (- t \pm 2)^{2} = 16$
$t^{2} + (t^{2} + 2 \pm 22 t) + (t^{2} + 2 \mp 22 t) = 16$
$3 t^{2} + 4 = 16 ⟹ 3 t^{2} = 12 ⟹ t = \pm 2$

Final Position Vectors of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

For $t = 2$ : $E = (1 + 2, 1 - 2, 1 - 2) = (3, - 1, - 1)$
For $t = - 2$ : $E = (1 - 2, 1 + 2, 1 + 2) = (- 1, 3, 3)$
Final position vectors: $(3, - 1, - 1)$ or $(- 1, 3, 3)$

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Visualized Solution (Hindi)

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Calculate Area of $△ A B C$

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Visualized Solution (Hindi)

Visualize the Tetrahedron .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABCD

Find the Median .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Calculate Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC

Find Altitude Height .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Define Coordinates of Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AD=4

Final Position Vectors of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

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Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + 2 \hat{j} - \hat{k}$ and $c = \hat{i} + \hat{j} - 2 \hat{k}$ be three vectors. A vector in the plane of $b$ and $c$ , whose projection on $a$ is of magnitude $2/3$ , is:

A tetrahedron has vertices at $O (0, 0, 0), A (1, 2, 1), B (2, 1, 3)$ and $C (- 1, 1, 2)$ . Then the angle between the faces OAB and ABC will be

Let $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = x \hat{i} + (x - 2) \hat{j} - \hat{k}$ . If the vectors $c$ lies in the plane of $a$ and $b$ , then $x$ equals

If $A = (1, 1, 1)$ , $C = (0, 1, - 1)$ are given vectors, then a vector $B$ satisfying the equations $A \times B = C$ and $A \cdot B = 3$ is .........