Let a vector a be coplanar with vectors b = 2i + j + k and c = i - j + k. If a is perpendicular to d

Q: Why is $[\vec{a} \vec{b} \vec{c}] = 0$?

The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ represents the volume of the parallelepiped formed by these three vectors. If $\vec{a}$ is coplanar with $\vec{b}$ and $\vec{c}$, the height of this parallelepiped is zero, so the volume must be zero. It is a geometric necessity!

Q: Why can we combine $[\vec{a} \vec{b} \vec{d}] + [\vec{a} \vec{c} \vec{d}]$ into $[\vec{a} (\vec{b} + \vec{c}) \vec{d}]$?

This is the linearity property of the scalar triple product. It behaves exactly like a determinant, and determinants are linear with respect to their rows. It is a powerful tool to simplify complex expressions and save you from tedious calculations.

Q: What if I get $\lambda = -1$ instead of $\lambda = 1$?

That is perfectly fine! The problem asks for 'a possible value'. Since the magnitude condition $|\vec{a}| = \sqrt{10}$ leads to $\lambda = \pm 1$, both values are valid. Choosing one will yield the same result for the final expression, as the vector $\vec{a}$ would just be in the opposite direction.

Question

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + \hat{k}$. If $\vec{a}$ is perpendicular to $\vec{d} = 3\hat{i} + 2\hat{j} + 6\hat{k}$, and $|\vec{a}| = \sqrt{10}$. Then a possible value of $[\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{d}] + [\vec{a} \vec{c} \vec{d}]$ is equal to:

Options:

(A) -42

(B) -40

(C) -29

(D) -38

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a vector problem; we are sculpting a solution in three-dimensional space.

Imagine you are standing in a coordinate system where vectors $\vec{b}$ and $\vec{c}$ define a flat, infinite plane. Your mystery vector, $\vec{a}$, is a resident of this plane.

If $\vec{a}$ is in the plane of $\vec{b}$ and $\vec{c}$, it must be a linear combination of them. We write:

$$\vec{a} = \lambda \vec{b} + \mu \vec{c}$$

This is the algebraic translation of the geometric fact that $\vec{a}$ is trapped within the span of $\vec{b}$ and $\vec{c}$.

### The Perpendicularity Constraint

Now, we are told $\vec{a}$ is perpendicular to $\vec{d}$. In the language of vectors, perpendicularity is synonymous with the dot product being zero:

$$\vec{a} \cdot \vec{d} = 0$$

By substituting our linear combination $\vec{a} = \lambda \vec{b} + \mu \vec{c}$ into this dot product, we create a bridge between the geometry and the algebra. We expand the expression:

$$(2\lambda + \mu)(3) + (\lambda - \mu)(2) + (\lambda + \mu)(6) = 0$$

As we simplify this, we find a beautiful relationship:

$$14\lambda + 7\mu = 0 \Rightarrow \mu = -2\lambda$$

We have successfully constrained our mystery vector to a single degree of freedom, $\lambda$.

### The Magnitude and the Final Shortcut

We are given $|\vec{a}| = \sqrt{10}$. This fixes the scale of our vector.

By substituting $\mu = -2\lambda$ back into our expression for $\vec{a}$, we find $\vec{a} = \lambda(0\hat{i} + 3\hat{j} - \hat{k})$. Applying the magnitude condition, we find $\lambda = \pm 1$.

Now, for the grand finale. We need to evaluate the sum of the scalar triple products:

$$[\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{d}] + [\vec{a} \vec{c} \vec{d}]$$

Since $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar, the first term $[\vec{a} \vec{b} \vec{c}]$ vanishes into thin air—it is zero. We are left with $[\vec{a} \vec{b} \vec{d}] + [\vec{a} \vec{c} \vec{d}]$.

Instead of calculating two separate determinants, we use the linearity property of the scalar triple product to combine them into:

$$[\vec{a} (\vec{b} + \vec{c}) \vec{d}]$$

This is the elegance of vector algebra. We calculate $\vec{b} + \vec{c} = 3\hat{i} + 0\hat{j} + 2\hat{k}$, set up our final determinant, and expand.

The result, **$-42$**, is not just a number; it is the reward for our logical journey. Keep practicing, keep visualizing, and remember that every vector has a story to tell.

Select Answer:

Visualized Solution

Visualizing the Vectors

The Coplanarity Condition

Representing $a$ Algebraically

Using Perpendicularity $a \cdot d = 0$

Solving for $μ$

Refining Vector $a$

Finding $λ$ using Magnitude

Simplifying the Expression

Calculating $b + c$

The Final Determinant

Final Answer

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Perpendicularity Constraint

The Magnitude and the Final Shortcut

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Vectors

The Coplanarity Condition

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

Using Perpendicularity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅d=0

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

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

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