What does the scalar $\lambda$ represent, and why does it matter?

The vector $\vec{a}$ is parallel to the intersection line, meaning $\vec{a}$ is a scalar multiple of the direction vector $\vec{v}$. We write this as $\vec{a} = \lambda \vec{v}$. The scalar $\lambda$ represents the magnitude and direction of $\vec{a}$. Because the problem asks for an obtuse angle, the sign of $\lambda$ becomes critical. If $\lambda$ is positive, the angle might be acute; if $\lambda$ is negative, it flips the vector $\vec{a}$, which is exactly what we need to satisfy the obtuse angle condition.

JEE Main 2022 (26 July Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Vector Algebra: A vector $a$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i} + \hat{j}$ and the plane determined by the vectors $\hat{i} - \hat{j}, \hat{i} + \hat{k}$ . The obtuse angle between $a$ and the vector $b = \hat{i} - 2 \hat{j} + 2 \hat{k}$ is

Q: Wait, why do we need the cross product of the normals? Can't we just solve the system of equations?

You absolutely could solve the system of linear equations, but that is often a longer path. Geometrically, the normal vector of a plane is perpendicular to every vector lying within that plane. Since the line of intersection lies on both planes, it must be perpendicular to both normal vectors. The cross product $\vec{n}_1 \times \vec{n}_2$ is the most direct way to find a vector perpendicular to both, giving you the direction of the intersection line instantly.

Q: How do I know if the angle is obtuse or acute just by looking at the dot product?

The dot product formula is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Since the magnitudes $|\vec{a}|$ and $|\vec{b}|$ are always positive, the sign of the dot product is determined entirely by $\cos \theta$. If the dot product is positive, $\cos \theta > 0$, meaning the angle is acute ($0 \leq \theta < \frac{\pi}{2}$). If the dot product is negative, $\cos \theta < 0$, meaning the angle is obtuse ($\frac{\pi}{2} < \theta \leq \pi$). By setting the dot product to be negative, we ensure our result is in the obtuse range.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the two planes in 3D space. The line of intersection is a unique line that lies on both planes. Geometrically, this means the direction of this line must be perpendicular to the normal vector of the first plane (

n_{1}

) and the normal vector of the second plane (

n_{2}

Hint 2

Hint 2: The Arsenal. To find a vector that is perpendicular to two given vectors, you must use the cross product. Calculate

n_{1}

by taking the cross product of the vectors spanning the first plane, and

n_{2}

by taking the cross product of the vectors spanning the second plane. Then, the direction vector of the intersection line is

v = n_{1} \times n_{2}

Hint 3

Hint 3: The Execution. Once you have the direction vector

v

, express

a

λ v

. Use the dot product formula

a \cdot b = ∣ a ∣∣ b ∣ cos θ

to find

cos θ

. Remember that the problem specifies an obtuse angle, which forces

cos θ < 0

, allowing you to determine the sign of the scalar

λ

Visualized Solution

Visualizing the Problem

Given: Vector $a$ is parallel to the line of intersection of two planes.
Plane 1 ( $P_{1}$ ) is spanned by $\hat{i}$ and $\hat{i} + \hat{j}$ .
Plane 2 ( $P_{2}$ ) is spanned by $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$ .
Target: Find the obtuse angle $θ$ between $a$ and $b = \hat{i} - 2 \hat{j} + 2 \hat{k}$ .

Finding Normal to Plane 1

A plane's normal vector is perpendicular to any two non-parallel vectors on it.
Normal to $P_{1}$ : $n_{1} = \hat{i} \times (\hat{i} + \hat{j})$
Using distributive property: $n_{1} = (\hat{i} \times \hat{i}) + (\hat{i} \times \hat{j})$
Since $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}$ :
$n_{1} = \hat{k}$

Finding Normal to Plane 2

Normal to $P_{2}$ : $n_{2} = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{k})$
Expanding the cross product:
$n_{2} = (\hat{i} \times \hat{i}) + (\hat{i} \times \hat{k}) - (\hat{j} \times \hat{i}) - (\hat{j} \times \hat{k})$
$n_{2} = 0 - \hat{j} + \hat{k} - \hat{i}$
$n_{2} = - \hat{i} - \hat{j} + \hat{k}$

Direction of Intersection Line

The line of intersection lies on both planes, so it is perpendicular to both $n_{1}$ and $n_{2}$ .
Direction vector $v = n_{1} \times n_{2}$
$v = \hat{k} \times (- \hat{i} - \hat{j} + \hat{k})$
$v = - (\hat{k} \times \hat{i}) - (\hat{k} \times \hat{j}) + (\hat{k} \times \hat{k})$
$v = - \hat{j} - (- \hat{i}) + 0 = \hat{i} - \hat{j}$

Defining Vector $a$

Since $a$ is parallel to the intersection line $v$ :
$a = λ (\hat{i} - \hat{j})$ for some scalar $λ \neq = 0$ .
We are given another vector $b = \hat{i} - 2 \hat{j} + 2 \hat{k}$ .
We need the angle between $a$ and $b$ .

Calculating the Dot Product

To find the angle, we use the dot product: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$
First, let's compute $a \cdot b$ :
$a \cdot b = λ (\hat{i} - \hat{j}) \cdot (\hat{i} - 2 \hat{j} + 2 \hat{k})$
$a \cdot b = λ [(1) (1) + (- 1) (- 2) + (0) (2)]$
$a \cdot b = λ [1 + 2 + 0] = 3 λ$

Magnitudes and Cosine Formula

Magnitude of $a$ : $∣ a ∣ = ∣ λ ∣ 1^{2} + (- 1)^{2} = 2 ∣ λ ∣$
Magnitude of $b$ : $∣ b ∣ = 1^{2} + (- 2)^{2} + 2^{2} = 1 + 4 + 4 = 9 = 3$
Substituting into the cosine formula:
$cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣} = \frac{3 λ}{( 2 ∣ λ ∣ ) ( 3 )}$
$cos θ = \frac{λ}{2 ∣ λ ∣}$

Solving for the Obtuse Angle

We are given that $θ$ is an obtuse angle.
For an obtuse angle ( $\frac{π}{2} < θ \leq π$ ), $cos θ < 0$ .
Therefore, $\frac{λ}{2 ∣ λ ∣} < 0$ , which implies $λ < 0$ .
If $λ < 0$ , then $∣ λ ∣ = - λ$ , so $\frac{λ}{∣ λ ∣} = - 1$ .
$cos θ = - \frac{1}{2}$

Final Conclusion

$cos θ = - \frac{1}{2}$
The principal value for $θ$ in $[0, π]$ is $π - \frac{π}{4} = \frac{3 π}{4}$ .
Final Answer: $θ = \frac{3 π}{4}$
Key Takeaway: The intersection of two planes is parallel to $n_{1} \times n_{2}$ . The sign of the scalar multiple is dictated by the angle constraints.

00:00 / 00:00

The Sigma Insight: Vector (Cross) Product

Solution Diagram

Analyzing the Setup

The problem asks us to find the obtuse angle between a vector

a

, which is parallel to the line of intersection of two planes, and a given vector

b = \hat{i} - 2 \hat{j} + 2 \hat{k}

The line of intersection of two planes is perpendicular to the normal vectors of both planes. Therefore, the direction vector

v

of the line of intersection is given by the cross product of the two normal vectors,

n_{1}

and

n_{2}

Phase 1

Unveiling the Normals

For the first plane

P_{1}

, spanned by

\hat{i}

and

\hat{i} + \hat{j}

, the normal vector

n_{1}

is:

n_{1} = \hat{i} \times (\hat{i} + \hat{j}) = \hat{k}

For the second plane

P_{2}

, spanned by

\hat{i} - \hat{j}

and

\hat{i} + \hat{k}

, the normal vector

n_{2}

is:

n_{2} = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{k}) = (\hat{i} \times \hat{i}) + (\hat{i} \times \hat{k}) - (\hat{j} \times \hat{i}) - (\hat{j} \times \hat{k})

Evaluating the cross products, we obtain:

n_{2} = 0 - \hat{j} - (- \hat{k}) - \hat{i} = - \hat{i} - \hat{j} + \hat{k}

Phase 2

The Line of Intersection

The direction vector

v

of the line of intersection is the cross product of the normals:

v = n_{1} \times n_{2} = \hat{k} \times (- \hat{i} - \hat{j} + \hat{k})

Distributing the cross product, we find:

v = - (\hat{k} \times \hat{i}) - (\hat{k} \times \hat{j}) + (\hat{k} \times \hat{k}) = - \hat{j} - (- \hat{i}) + 0 = \hat{i} - \hat{j}

Phase 3

The Angle of Engagement

Since

a

is parallel to

v

, we have

a = λ (\hat{i} - \hat{j})

for some scalar $\lambda eq 0$. We use the dot product formula

a \cdot b = ∣ a ∣∣ b ∣ cos θ

to find the angle

θ

The dot product is:

a \cdot b = λ (\hat{i} - \hat{j}) \cdot (\hat{i} - 2 \hat{j} + 2 \hat{k}) = λ (1 + 2 + 0) = 3 λ

The magnitudes are:

∣ a ∣ = ∣ λ ∣ 1^{2} + (- 1)^{2} = ∣ λ ∣ 2, ∣ b ∣ = 1^{2} + (- 2)^{2} + 2^{2} = 3

Substituting these into the cosine formula:

cos θ = \frac{3 λ}{3∣ λ ∣ 2} = \frac{λ}{∣ λ ∣ 2}

Since the problem specifies an obtuse angle,

cos θ

must be negative, which implies

λ < 0

. Thus,

\frac{λ}{∣ λ ∣} = - 1

, leading to:

cos θ = - \frac{1}{2}

The angle

θ

in the range

[0, π]

satisfying this condition is $\frac{3 π}{4}$ .

Common Traps & FAQs

Wait, why do we need the cross product of the normals? Can't we just solve the system of equations?

You absolutely could solve the system of linear equations, but that is often a longer path. Geometrically, the normal vector of a plane is perpendicular to every vector lying within that plane. Since the line of intersection lies on both planes, it must be perpendicular to both normal vectors. The cross product

n_{1} \times n_{2}

is the most direct way to find a vector perpendicular to both, giving you the direction of the intersection line instantly.

What does the scalar

λ

represent, and why does it matter?

The vector

a

is parallel to the intersection line, meaning

a

is a scalar multiple of the direction vector

v

. We write this as

a = λ v

. The scalar

λ

represents the magnitude and direction of

a

. Because the problem asks for an obtuse angle, the sign of

λ

becomes critical. If

λ

is positive, the angle might be acute; if

λ

is negative, it flips the vector

a

, which is exactly what we need to satisfy the obtuse angle condition.

How do I know if the angle is obtuse or acute just by looking at the dot product?

The dot product formula is

a \cdot b = ∣ a ∣∣ b ∣ cos θ

. Since the magnitudes

∣ a ∣

and

∣ b ∣

are always positive, the sign of the dot product is determined entirely by

cos θ

. If the dot product is positive,

cos θ > 0

, meaning the angle is acute (

0 \leq θ < \frac{π}{2}

). If the dot product is negative,

cos θ < 0

, meaning the angle is obtuse (

\frac{π}{2} < θ \leq π

). By setting the dot product to be negative, we ensure our result is in the obtuse range.

Select Answer:

Visualized Solution

Visualizing the Problem

Finding Normal to Plane 1

Finding Normal to Plane 2

Direction of Intersection Line

Defining Vector $a$

Calculating the Dot Product

Magnitudes and Cosine Formula

Solving for the Obtuse Angle

Final Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

Let the vectors $a, b, c$ and $d$ be such that $(a \times b) \times (c \times d) = 0$ . Let $P_{1}$ and $P_{2}$ be planes determined by the pairs of vectors $a, b$ and $c, d$ respectively. Then the angle between $P_{1}$ and $P_{2}$ is

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $a \cdot c = ∣ c ∣, ∣ c - a ∣ = 22$ and the angle between $(a \times b)$ and $c$ is $3 0^{\circ}$ , then $∣ (a \times b) \times c ∣ =$

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c = a \times b$ . Let $d$ be a vector such that $∣ d - a ∣ = 11, ∣ c \times d ∣ = 3$ and the angle between $c$ and $d$ is $\frac{π}{4}$ . Then $a \cdot d$ is equal to

Let $a$ and $b$ be the vectors along the diagonal of a parallelogram having area $22$ . Let the angle between $a$ and $b$ be acute. $∣ a ∣ = 1$ and $∣ a \cdot b ∣ = ∣ a \times b ∣$ . If $c = 22 (a \times b) - 2 b$ , then an angle between $b$ and $c$ is:

Let $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ be two vectors such that $∣ a ∣ = 1$ ; $a \cdot b = 2$ and $∣ b ∣ = 4$ . If $c = 2 (a \times b) - 3 b$ , then the angle between $b$ and $c$ is equal to :

Let $a, b, c$ be three vectors such that $∣ a ∣ = 31, 4∣ b ∣ = ∣ c ∣ = 2$ and $2 (a \times b) = 3 (c \times a)$ . If the angle between $b$ and $c$ is $2 π /3$ , then $(\frac{a \times c}{a \cdot b})^{2}$ is equal to

Let $p = 2 \hat{i} + 3 \hat{j} + \hat{k}$ and $q = \hat{i} + 2 \hat{j} + \hat{k}$ be two vectors. If a vector $r = (α \hat{i} + β \hat{j} + γ \hat{k})$ is perpendicular to each of the vectors $(p + q)$ and $(p - q)$ , and $∣ r ∣ = 3$ , then $∣ α ∣ + ∣ β ∣ + ∣ γ ∣$ is equal to

Let $\overset{a}{^}, \hat{b}$ be unit vectors. If $c$ be a vector such that the angle between $\overset{a}{^}$ and $c$ is $\frac{π}{12}$ , and $\hat{b} = c + 2 (c \times \overset{a}{^})$ , then $∣6 c ∣^{2}$ is equal to

Let $α = 3 \hat{i} + \hat{j}$ and $β = 2 \hat{i} - \hat{j} + 3 \hat{k}$ . If $β = β_{1} - β_{2}$ , where $β_{1}$ is parallel to $α$ and $β_{2}$ is perpendicular to $α$ , then $β_{1} \times β_{2}$ is equal to

Select Answer:

Visualized Solution

Visualizing the Problem

Finding Normal to Plane 1

Finding Normal to Plane 2

Direction of Intersection Line

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

Calculating the Dot Product

Magnitudes and Cosine Formula

Solving for the Obtuse Angle

Final Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+j^​−2k^ and b=i^+j^​. If c is a vector such that a⋅c=∣c∣,∣c−a∣=22​ and the angle between (a×b) and c is 30∘, then ∣(a×b)×c∣=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+j^​−2k^,b=i^+j^​ and c=a×b. Let d be a vector such that ∣d−a∣=11​,∣c×d∣=3 and the angle between c and d is 4π​. Then a⋅d is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c be three vectors such that ∣a∣=31​,4∣b∣=∣c∣=2 and 2(a×b)=3(c×a). If the angle between b and c is 2π/3, then (a⋅ba×c​)2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a^,b^ be unit vectors. If c be a vector such that the angle between a^ and c is 12π​, and b^=c+2(c×a^), then ∣6c∣2 is equal to

Defining Vector $a$

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $a \cdot c = ∣ c ∣, ∣ c - a ∣ = 22$ and the angle between $(a \times b)$ and $c$ is $3 0^{\circ}$ , then $∣ (a \times b) \times c ∣ =$

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c = a \times b$ . Let $d$ be a vector such that $∣ d - a ∣ = 11, ∣ c \times d ∣ = 3$ and the angle between $c$ and $d$ is $\frac{π}{4}$ . Then $a \cdot d$ is equal to

Let $a, b, c$ be three vectors such that $∣ a ∣ = 31, 4∣ b ∣ = ∣ c ∣ = 2$ and $2 (a \times b) = 3 (c \times a)$ . If the angle between $b$ and $c$ is $2 π /3$ , then $(\frac{a \times c}{a \cdot b})^{2}$ is equal to

Let $\overset{a}{^}, \hat{b}$ be unit vectors. If $c$ be a vector such that the angle between $\overset{a}{^}$ and $c$ is $\frac{π}{12}$ , and $\hat{b} = c + 2 (c \times \overset{a}{^})$ , then $∣6 c ∣^{2}$ is equal to