JEE Advanced 2004S

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: The unit vector which is orthogonal to the vector $3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and is coplanar with the vectors $2 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} - \hat{j} + \hat{k}$ is

Q: Why do we use $\lambda$ in $\vec{r} = \vec{v} + \lambda\vec{w}$? Can't we use two scalars like $\alpha\vec{v} + \beta\vec{w}$?

You absolutely could! Using $\alpha\vec{v} + \beta\vec{w}$ is perfectly valid. However, since we are looking for a *direction* (a unit vector), the ratio of the coefficients matters more than their absolute values. By setting one coefficient to 1 (or using a single parameter $\lambda$), we simplify the algebra significantly without losing any generality.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. If a vector is coplanar with two others, it must reside in the plane spanned by them. This means it can be expressed as a linear combination of those two vectors, like a recipe using two base ingredients. Think of this as finding a vector that is a 'weighted sum' of the two given vectors.

Hint 2

Hint 2: Recall the 'Toolbox of Orthogonality.' When two vectors are perpendicular, their dot product is zero. You have a vector

r

that is a linear combination of

v

and

w

, and it must satisfy the condition

r \cdot u = 0

. This is the bridge between the geometric requirement and the algebraic solution.

Hint 3

Hint 3: Set up the equation: Let

r = v + λ w

. Substitute the components to get

r = (2 + λ) \hat{i} + (1 - λ) \hat{j} + (1 + λ) \hat{k}

. Now, compute the dot product with

u = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}

and set it to zero to solve for the scalar

λ

Visualized Solution

Defining the Vectors

Let $u = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$
Let $v = 2 \hat{i} + \hat{j} + \hat{k}$
Let $w = \hat{i} - \hat{j} + \hat{k}$

The Coplanarity Condition

A vector $r$ coplanar with $v$ and $w$ is given by:
$r = v + λ w$

Substituting Components

$r = (2 \hat{i} + \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$

Grouping Components

$r = (2 + λ) \hat{i} + (1 - λ) \hat{j} + (1 + λ) \hat{k}$

Applying Orthogonality

Since $r ⊥ u$ , their dot product is zero:
$r \cdot u = 0$

Setting up the Dot Product

$u = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$
$r \cdot u = 3 (2 + λ) + 2 (1 - λ) + 6 (1 + λ) = 0$

Expanding the Equation

$6 + 3 λ + 2 - 2 λ + 6 + 6 λ = 0$

Solving for $λ$

$7 λ + 14 = 0$
$7 λ = - 14$
$λ = - 2$

Finding Vector $r$

Substitute $λ = - 2$ back into $r$ :
$r = (2 + (- 2)) \hat{i} + (1 - (- 2)) \hat{j} + (1 + (- 2)) \hat{k}$

Calculating Vector $r$

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

Normalizing the Vector

The question asks for a unit vector.
$\overset{r}{^} = \pm \frac{r}{∣ r ∣}$

Calculating Magnitude

$∣ r ∣ = 0^{2} + 3^{2} + (- 1)^{2}$
$∣ r ∣ = 0 + 9 + 1$
$∣ r ∣ = 10$

Final Unit Vector

Unit vector $= \pm \frac{3 j ^ - k ^}{10}$
Comparing with options, the correct choice is $\frac{3 j ^ - k ^}{10}$ (assuming a typo in option C where $\hat{i}$ should be $\hat{j}$ ).

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The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Setup

We are given two vectors,

v = 2 \hat{i} + \hat{j} + \hat{k}

and

w = \hat{i} - \hat{j} + \hat{k}

, which define a plane in three-dimensional space. We seek a vector

r

that lies within this plane and is simultaneously orthogonal to a third vector,

u = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}

The Geometry of Coplanarity

If a vector

r

is coplanar with

v

and

w

, it must be a linear combination of these two vectors. We express this relationship as:

r = v + λ w

Substituting the given components into this expression, we have:

r = (2 \hat{i} + \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})

Grouping the terms by their respective unit vectors, we obtain the general form for any vector in the plane:

r = (2 + λ) \hat{i} + (1 - λ) \hat{j} + (1 + λ) \hat{k}

The Orthogonality Constraint

The problem imposes the condition that

r

must be orthogonal to

u

. In vector algebra, this implies that their dot product must be zero:

r \cdot u = 0

Substituting our expression for

r

and the given vector

u = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}

, we set up the following equation:

3 (2 + λ) + 2 (1 - λ) + 6 (1 + λ) = 0

Expanding the terms, we get:

6 + 3 λ + 2 - 2 λ + 6 + 6 λ = 0

Combining like terms results in:

7 λ + 14 = 0 \Rightarrow λ = - 2

Final Calculation

Substituting

λ = - 2

back into our general expression for

r

, we find:

r = (2 - 2) \hat{i} + (1 - (- 2)) \hat{j} + (1 - 2) \hat{k}

r = 3 \hat{j} - \hat{k}

To find the unit vector, we first calculate the magnitude of

r

∣ r ∣ = 0^{2} + 3^{2} + (- 1)^{2} = 9 + 1 = 10

Normalizing the vector, we arrive at the final result:

\hat{r} = \pm \frac{3 j ^ - k ^}{10}

Common Traps & FAQs

Why can't we just take the cross product of the two vectors to find the coplanar one?

That is a brilliant question! The cross product of two vectors gives you a vector that is *perpendicular* to the plane containing them, not *in* the plane. We need a vector that lives inside the plane, so we use a linear combination, which allows us to span any direction within that flat surface.

Why do we use

λ

r = v + λ w

? Can't we use two scalars like

α v + β w

You absolutely could! Using

α v + β w

is perfectly valid. However, since we are looking for a *direction* (a unit vector), the ratio of the coefficients matters more than their absolute values. By setting one coefficient to 1 (or using a single parameter

λ

), we simplify the algebra significantly without losing any generality.

Do I really need to normalize at the end? Why is the magnitude important?

Yes, you must! The question explicitly asks for a *unit* vector. A vector with a magnitude of

10

is not a unit vector. Normalization is the final, crucial step that scales your vector down to a length of 1, ensuring your answer meets the strict definition required by the problem.

Animated Solution for Mathematics - Vector Algebra: The unit vector which is orthogonal to the vector $3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and is coplanar with the vectors $2 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} - \hat{j} + \hat{k}$ is

Select Answer:

Visualized Solution

Defining the Vectors

The Coplanarity Condition

Substituting Components

Grouping Components

Applying Orthogonality

Setting up the Dot Product

Expanding the Equation

Solving for $λ$

Finding Vector $r$

Calculating Vector $r$

Normalizing the Vector

Calculating Magnitude

Final Unit Vector

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Geometry of Coplanarity

The Orthogonality Constraint

Final Calculation

Similar Questions

The vector(s) which is/are coplanar with vectors $\hat{i} + \hat{j} + 2 \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ , and perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$ is/are

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

If $a$ is nonzero vector such that its projections on the vectors $2 \hat{i} - \hat{j} + 2 \hat{k}, \hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $a$ is:

Let $u$ be a vector coplanar with the vectors $a = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $b = \hat{j} + \hat{k}$ . If $u$ is perpendicular to $a$ and $u \cdot b = 24$ , then $∣ u ∣^{2}$ is equal to :

Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} + \hat{j} - \hat{k}$ . A vector in the plane of $a$ and $b$ whose projection on $c$ is $\frac{1}{3}$ , is

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three vectors. A vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{3}$ , is given by

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:

Select Answer:

Visualized Solution

Defining the Vectors

The Coplanarity Condition

Substituting Components

Grouping Components

Applying Orthogonality

Setting up the Dot Product

Expanding the Equation

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

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

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

Normalizing the Vector

Calculating Magnitude

Final Unit Vector

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Geometry of Coplanarity

The Orthogonality Constraint

Final Calculation

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Solving for $λ$

Finding Vector $r$

Calculating Vector $r$

The vector(s) which is/are coplanar with vectors $\hat{i} + \hat{j} + 2 \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ , and perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$ is/are

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

If $a$ is nonzero vector such that its projections on the vectors $2 \hat{i} - \hat{j} + 2 \hat{k}, \hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $a$ is:

Let $u$ be a vector coplanar with the vectors $a = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $b = \hat{j} + \hat{k}$ . If $u$ is perpendicular to $a$ and $u \cdot b = 24$ , then $∣ u ∣^{2}$ is equal to :

Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} + \hat{j} - \hat{k}$ . A vector in the plane of $a$ and $b$ whose projection on $c$ is $\frac{1}{3}$ , is

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