JEE Main 2023 (30 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: If a plane passes through the points $(- 1, k, 0)$ , $(2, k, - 1)$ , $(1, 1, 2)$ and is parallel to the line $\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k ^{2} + 1}{( k - 1 ) ( k - 2 )}$ is

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Visualized Solution

Visualizing the Geometry

Points on the plane: $A (- 1, k, 0)$ , $B (2, k, - 1)$ , and $C (1, 1, 2)$
Line equation: $\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$
Goal: Find $k$ and evaluate $\frac{k ^{2} + 1}{( k - 1 ) ( k - 2 )}$

Vectors on the Plane

To define the plane, we need two vectors lying on it.
We construct vectors $A B$ and $A C$ using the given points.

Calculating Vector $A B$

$A B = (2 - (- 1)) \hat{i} + (k - k) \hat{j} + (- 1 - 0) \hat{k}$
$A B = 3 \hat{i} + 0 \hat{j} - 1 \hat{k}$

Calculating Vector $A C$

$A C = (1 - (- 1)) \hat{i} + (1 - k) \hat{j} + (2 - 0) \hat{k}$
$A C = 2 \hat{i} + (1 - k) \hat{j} + 2 \hat{k}$

The Normal Vector $n$

The normal vector $n$ is perpendicular to the plane.
$n = A B \times A C$

Evaluating $A B \times A C$

$n = \hat{i} 32 \hat{j} 0 1 - k \hat{k} - 1 2$
$n = \hat{i} (0 - (- 1) (1 - k)) - \hat{j} (6 - (- 2)) + \hat{k} (3 (1 - k) - 0)$
$n = (1 - k) \hat{i} - 8 \hat{j} + (3 - 3 k) \hat{k}$

Direction Vector of Line $L$

Given line: $\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$
Caution: The $y$ -term is not in standard form!
Rewrite as: $\frac{x - 1}{1} = \frac{y + 1/2}{1} = \frac{z + 1}{- 1}$
Direction vector $d = 1 \hat{i} + 1 \hat{j} - 1 \hat{k}$

Applying the Parallel Condition

The plane is parallel to the line $L$ .
Therefore, $n ⊥ d$
Condition for perpendicularity: $n \cdot d = 0$

Setting up the Dot Product

$n = (1 - k) \hat{i} - 8 \hat{j} + (3 - 3 k) \hat{k}$
$d = 1 \hat{i} + 1 \hat{j} - 1 \hat{k}$
$n \cdot d = (1 - k) (1) + (- 8) (1) + (3 - 3 k) (- 1) = 0$

Solving for $k$

Expand the equation: $1 - k - 8 - 3 + 3 k = 0$
Combine like terms: $2 k - 10 = 0$
$2 k = 10 \Rightarrow k = 5$

Final Evaluation

Expression to evaluate: $\frac{k ^{2} + 1}{( k - 1 ) ( k - 2 )}$
Substitute $k = 5$ : $\frac{5 ^{2} + 1}{( 5 - 1 ) ( 5 - 2 )}$
$= \frac{25 + 1}{( 4 ) ( 3 )} = \frac{26}{12}$
Final Answer: $\frac{13}{6}$

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

To define the plane, we identify two vectors lying on its surface by anchoring them at point

A (- 1, k, 0)

. We use points

B (2, k, - 1)

and

C (1, 1, 2)

to construct these vectors.

The vector

A B

is calculated as:

A B = (2 - (- 1)) \hat{i} + (k - k) \hat{j} + (- 1 - 0) \hat{k} = 3 \hat{i} + 0 \hat{j} - 1 \hat{k}

The vector

A C

is calculated as:

A C = (1 - (- 1)) \hat{i} + (1 - k) \hat{j} + (2 - 0) \hat{k} = 2 \hat{i} + (1 - k) \hat{j} + 2 \hat{k}

The Normal Vector

The normal vector

n

is perpendicular to the plane and is found by the cross product

n = A B \times A C

n = \hat{i} 32 \hat{j} 0 1 - k \hat{k} - 1 2

Expanding the determinant, we obtain:

n = \hat{i} (0 - (- 1) (1 - k)) - \hat{j} (6 - (- 2)) + \hat{k} (3 (1 - k) - 0)

Simplifying the components, the normal vector is:

n = (1 - k) \hat{i} - 8 \hat{j} + (3 - 3 k) \hat{k}

The Line Trap

The line is given by the equation

\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}

. To extract the direction vector

d

, we must ensure the coefficient of

y

1

Rewriting the middle term as

\frac{y + 1/2}{1}

, we identify the direction vector as:

d = 1 \hat{i} + 1 \hat{j} - 1 \hat{k}

The Parallel Condition

Since the plane is parallel to the line, the normal vector

n

must be perpendicular to the direction vector

d

. Therefore, their dot product must be zero:

n \cdot d = 0

Substituting the vectors into the dot product equation:

(1 - k) (1) + (- 8) (1) + (3 - 3 k) (- 1) = 0

Expanding and solving for

k

1 - k - 8 - 3 + 3 k = 0

2 k - 10 = 0

k = 5

Final Calculation

With the value

k = 5

determined, we evaluate the target expression:

\frac{k ^{2} + 1}{( k - 1 ) ( k - 2 )}

Substituting

k = 5

\frac{5 ^{2} + 1}{( 5 - 1 ) ( 5 - 2 )} = \frac{25 + 1}{4 \times 3} = \frac{26}{12}

Reducing the fraction, we arrive at the final result:

\frac{13}{6}

Common Traps & FAQs

Wait, why do we need the cross product? Can't we just use the points directly?

The cross product is our bridge between the 2D surface of the plane and the 3D space around it. It generates the normal vector, which is the 'DNA' of the plane. Without it, we cannot define the orientation of the plane relative to the line. It is the most elegant way to capture the tilt of the plane in space.

Why is the line equation a trap? It looks correct.

In JEE Advanced, examiners love to test your attention to detail. The term

\frac{2 y + 1}{2}

is not in the standard form

\frac{y - y _{1}}{b}

. The coefficient of

y

must be

1

. By dividing the numerator and denominator by

2

, we reveal the true direction vector. If you miss this, your direction vector will be wrong, and the entire calculation will collapse.

Why does the dot product equal zero?

If the line is parallel to the plane, it never intersects it. This implies the line's direction vector is perpendicular to the plane's normal vector. The dot product of two perpendicular vectors is always zero—a fundamental property of orthogonality. It is the mathematical manifestation of the line 'gliding' alongside the plane.

Animated Solution for Mathematics - Three Dimensional Geometry: If a plane passes through the points $(- 1, k, 0)$ , $(2, k, - 1)$ , $(1, 1, 2)$ and is parallel to the line $\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k ^{2} + 1}{( k - 1 ) ( k - 2 )}$ is

Select Answer:

Visualized Solution

Visualizing the Geometry

Vectors on the Plane

Calculating Vector $A B$

Calculating Vector $A C$

The Normal Vector $n$

Evaluating $A B \times A C$

Direction Vector of Line $L$

Applying the Parallel Condition

Setting up the Dot Product

Solving for $k$

Final Evaluation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Normal Vector

The Line Trap

The Parallel Condition

Final Calculation

Similar Questions

Let $(λ, 2, 1)$ be a point on the plane which passes through the point $(4, - 2, 2)$ . If the plane is perpendicular to the line joining the point $(- 2, - 21, 29)$ and $(- 1, - 16, 23)$ , then $(\frac{λ}{11})^{2} - \frac{4 λ}{11} - 4$ is equal to

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

If the equation of the plane passing through the point $(1, 1, 2)$ and perpendicular to the line $x - 3 y + 2 z - 1 = 0 = 4 x - y + z$ is $A x + B y + C z = 1$ , then $140 (C - B + A)$ is equal to ______.

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

A plane $E$ is perpendicular to the two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $32$ , then $(P Q)^{2}$ is equal to

If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3, 4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + cz + 6 = 0$ , then $a + b + c$ is equal to

A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

Let $P$ be a plane containing the line $\frac{x - 1}{3} = \frac{y + 6}{4} = \frac{z + 5}{2}$ and parallel to the line $\frac{x - 3}{4} = \frac{y - 2}{- 3} = \frac{z + 5}{7}$ . If the point $(1, - 1, α)$ lies on the plane $P$ , then the value of $∣5 α ∣$ is equal to ___

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Visualized Solution

Visualizing the Geometry

Vectors on the Plane

Calculating Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Calculating Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

The Normal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB×AC

Direction Vector of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Applying the Parallel Condition

Setting up the Dot Product

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Final Evaluation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Normal Vector

The Line Trap

The Parallel Condition

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x−3y+2z−1=0=4x−y+z is Ax+By+Cz=1, then 140(C−B+A) is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane passing through the point (4, -1, 2) and parallel to the lines 3x+2​=−1y−2​=2z+1​ and 1x−2​=2y−3​=3z−4​ also passes through the point :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane passing through the points (0,−1,0) and (0,0,1) and making an angle π/4 with the plane y−z+5=0, also passes through the point

Calculating Vector $A B$

Calculating Vector $A C$

The Normal Vector $n$

Evaluating $A B \times A C$

Direction Vector of Line $L$

Solving for $k$

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

If the equation of the plane passing through the point $(1, 1, 2)$ and perpendicular to the line $x - 3 y + 2 z - 1 = 0 = 4 x - y + z$ is $A x + B y + C z = 1$ , then $140 (C - B + A)$ is equal to ______.

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point