Let λ ∈ Z, a = λi + j - k and b = 3i - j + 2k. Let c be a vector such that (a + b + c) × c = 0, a ·

Q: Why can we assume \vec{c} = k(\vec{a} + \vec{b})? What if \vec{c} is the zero vector?

That is an excellent conceptual check! If \vec{c} were the zero vector, then \vec{a} \cdot \vec{c} would be 0, not -17. Since the problem gives us non-zero dot products, \vec{c} cannot be the zero vector. Therefore, the cross product condition implies that \vec{c} and (\vec{a} + \vec{b}) are parallel, which mathematically necessitates the scalar multiple k.

Q: Why divide the equations? Can't we just solve for k first?

You absolutely could! But in the heat of a JEE Advanced exam, efficiency is your best friend. Dividing the equations eliminates k instantly, saving you from solving a system of linear equations and letting you jump straight to the quadratic in \lambda. It is a strategic shortcut that minimizes calculation errors.

Q: Why do we reject the non-integer root for \lambda?

The problem explicitly states \lambda \in \mathbb{Z}. In JEE Advanced, constraints are not suggestions; they are filters. If a value violates the domain (in this case, the set of integers), it is mathematically invalid for the context of the problem. Always check your constraints at the end!

Question

Let $\lambda \in \mathbb{Z}$, $\vec{a} = \lambda\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a} + \vec{b} + \vec{c}) 	imes \vec{c} = \vec{0}$, $\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Then $|\vec{c} 	imes (\lambda\hat{i} + \hat{j} + \hat{k})|^2$ is equal to

Options:

(A) 46

(B) 53

(C) 62

(D) 49

Sigma Percentile · Accepted Answer

### Analyzing the Vector Geometry

The problem begins with the vector equation:
$$(\vec{a} + \vec{b} + \vec{c}) 	imes \vec{c} = \vec{0}$$

We utilize the distributive property of the cross product to expand this expression:
$$(\vec{a} + \vec{b}) 	imes \vec{c} + \vec{c} 	imes \vec{c} = \vec{0}$$

Since the cross product of any vector with itself is the zero vector, specifically $\vec{c} 	imes \vec{c} = \vec{0}$, the equation simplifies to:
$$(\vec{a} + \vec{b}) 	imes \vec{c} = \vec{0}$$

### Establishing Collinearity

The condition $(\vec{a} + \vec{b}) 	imes \vec{c} = \vec{0}$ implies that the vector $\vec{c}$ is collinear with the vector sum $(\vec{a} + \vec{b})$. We can express this relationship using a scalar constant $k$:
$$\vec{c} = k(\vec{a} + \vec{b})$$

Given $\vec{a} = \lambda\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$, we calculate the sum:
$$\vec{a} + \vec{b} = (\lambda + 3)\hat{i} + 0\hat{j} + 1\hat{k}$$

Thus, our vector $\vec{c}$ is defined as:
$$\vec{c} = k((\lambda + 3)\hat{i} + \hat{k})$$

### Solving for Parameters

We are provided with the dot product conditions $\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Substituting $\vec{c}$ into these equations:
$$k[\lambda(\lambda + 3) - 1] = -17$$
$$k[3(\lambda + 3) + 2] = -20$$

To eliminate $k$, we divide the first equation by the second:
$$\frac{\lambda^2 + 3\lambda - 1}{3\lambda + 11} = \frac{17}{20}$$

Cross-multiplying yields the quadratic equation:
$$20\lambda^2 + 60\lambda - 20 = 51\lambda + 187$$
$$20\lambda^2 + 9\lambda - 207 = 0$$

Solving for $\lambda$ using the quadratic formula, we find the roots. Given the constraint $\lambda \in \mathbb{Z}$, we identify the valid integer solution:
$$\lambda = 3$$

### Final Calculation

Substituting $\lambda = 3$ back into our equations, we find $k = -1$. Consequently, the vector $\vec{c}$ is:
$$\vec{c} = -6\hat{i} - \hat{k}$$

We now compute the cross product of $\vec{c}$ and $(3\hat{i} + \hat{j} + \hat{k})$:
$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -6 & 0 & -1 \ 3 & 1 & 1 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(-6 - (-3)) + \hat{k}(-6 - 0)$$
$$\vec{v} = \hat{i} + 3\hat{j} - 6\hat{k}$$

The magnitude squared of this vector is:
$$|\vec{v}|^2 = 1^2 + 3^2 + (-6)^2 = 1 + 9 + 36 = 46$$

The final result is **46**.

Select Answer:

Visualized Solution

Analyze the Cross Product Condition

The Concept of Collinearity

Calculate the Sum Vector $a + b$

Using the First Dot Product Condition

Using the Second Dot Product Condition

Eliminating the Scalar $k$

Solving the Quadratic Equation

Finding the Integer Value of $λ$

Finding the Scalar $k$ and Vector $c$

Setting up the Final Cross Product

Evaluating the Determinant

Final Calculation: Magnitude Squared

The Sigma Insight: Vector (Cross) Product

Analyzing the Vector Geometry

Establishing Collinearity

Solving for Parameters

Final Calculation

Similar Questions

Let $a = 2 \hat{i} - 5 \hat{j} + 5 \hat{k}$ and $b = \hat{i} - \hat{j} + 3 \hat{k}$ . If $c$ is a vector such that $2 (a \times c) + 3 (b \times c) = 0$ and $(a - b) \cdot c = - 97$ , then $∣ c \times \hat{k} ∣^{2}$ is equal to

Let $a = 3 \hat{i} + \hat{j} - \hat{k}$ and $c = 2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ . If $b$ is a vector such that $a = b \times c$ and $∣ b ∣^{2} = 50$ , then $∣72 - ∣ b + c ∣^{2} ∣$ is equal to

Let $a = \hat{i} + 2 \hat{j} + λ \hat{k}$ , $b = 3 \hat{i} - 5 \hat{j} - λ \hat{k}$ , $a \cdot c = 7$ , $2 b \cdot c + 43 = 0$ , $a \times c = b \times c$ . Then $∣ a \cdot b ∣$ is equal to

Let $a = \hat{j} - \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ . Then the vector $b$ satisfying $a \times b + c = 0$ and $a \cdot b = 3$

Let $a = - \hat{i} + 2 \hat{j} + 2 \hat{k}, b = 8 \hat{i} + 7 \hat{j} - 3 \hat{k}$ and $c$ be vector such that $a \times c = b$ . If $c \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ , then $∣ a + c ∣^{2}$ is equal to :

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = 5 \hat{i} - 3 \hat{j} + 3 \hat{k}$ be there vectors. If $r$ is a vector such that, $r \times b = c \times b$ and $r \cdot a = 0$ . Then $25∣ r ∣^{2}$ is equal to

Let $a = 4 \hat{i} - \hat{j} + \hat{k}, b = 11 \hat{i} - \hat{j} + \hat{k}$ and $c$ be a vector such that $(a + b) \times c = c \times (- 2 a + 3 b)$ . If $(2 a + 3 b) \cdot c = 1670$ , then $∣ c ∣^{2}$ is equal to :

Let $a = 2 \hat{i} - 3 \hat{j} + \hat{k}, b = 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$ and a vector $c$ be such that $(a - c) \times b = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}$ and $a \cdot c = 3$ . If $b \times c = d$ , then $∣ a \cdot d ∣$ is equal to :

If $a = \hat{i} + 2 \hat{k}, b = \hat{i} + \hat{j} + \hat{k}, c = 7 \hat{i} - 3 \hat{j} + 4 \hat{k}$ , $r \times b + b \times c = 0$ and $r \cdot a = 0$ then $r \cdot c$ is equal to

Let $a = \hat{i} + 4 \hat{j} + 2 \hat{k}$ , $b = 3 \hat{i} - \hat{j} + 7 \hat{k}$ and $c = 2 \hat{i} - \hat{j} + 4 \hat{k}$ . If a vector $d$ satisfies $d \times b = c \times b$ and $d \cdot a = 24$ , then $∣ d ∣^{2}$ is equal to

Select Answer:

Visualized Solution

Analyze the Cross Product Condition

The Concept of Collinearity

Calculate the Sum Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b

Using the First Dot Product Condition

Using the Second Dot Product Condition

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

Solving the Quadratic Equation

Finding the Integer Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the Scalar .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k and Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Setting up the Final Cross Product

Evaluating the Determinant

Final Calculation: Magnitude Squared

The Sigma Insight: Vector (Cross) Product

Analyzing the Vector Geometry

Establishing Collinearity

Solving for Parameters

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^−5j^​+5k^ and b=i^−j^​+3k^. If c is a vector such that 2(a×c)+3(b×c)=0 and (a−b)⋅c=−97, then ∣c×k^∣2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=3i^+j^​−k^ and c=2i^−3j^​+3k^. If b is a vector such that a=b×c and ∣b∣2=50, then ∣72−∣b+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=i^+2j^​+λk^, b=3i^−5j^​−λk^, a⋅c=7, 2b⋅c+43=0, a×c=b×c. Then ∣a⋅b∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=j^​−k^ and c=i^−j^​−k^. Then the vector b satisfying a×b+c=0 and a⋅b=3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=−i^+2j^​+2k^,b=8i^+7j^​−3k^ and c be vector such that a×c=b. If c⋅(i^+j^​+k^)=4, then ∣a+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=i^+2j^​+3k^,b=i^−j^​+2k^ and c=5i^−3j^​+3k^ be there vectors. If r is a vector such that, r×b=c×b and r⋅a=0. Then 25∣r∣2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=4i^−j^​+k^,b=11i^−j^​+k^ and c be a vector such that (a+b)×c=c×(−2a+3b). If (2a+3b)⋅c=1670, then ∣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=2i^−3j^​+k^,b=3i^+2j^​+5k^ and a vector c be such that (a−c)×b=−18i^−3j^​+12k^ and a⋅c=3. If b×c=d, then ∣a⋅d∣ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=i^+2k^,b=i^+j^​+k^,c=7i^−3j^​+4k^, r×b+b×c=0 and r⋅a=0 then r⋅c is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=i^+4j^​+2k^, b=3i^−j^​+7k^ and c=2i^−j^​+4k^. If a vector d satisfies d×b=c×b and d⋅a=24, then ∣d∣2 is equal to