Let a = 2 i + j + k, and b and c be two nonzero vectors such that | a + b + c | = | a + b - c | and b c = 0. Consider the following two statements: (A) The magnitude of the vector a plus times the…

Given: a = 2 i + j + k, | a + b + c | = | a + b - c |, and b c = 0. Find: Which of the statements (A) and (B) is correct. Start with | a + b + c | = | a + b - c |. Squaring both sides, ( a + b + c)^2 = ( a + b - c)^2. Expanding, a^2 + 2 a b + 2 a c + b^2 + 2 b c + c^2 = a^2 + 2 a b - 2 a c + b^2 - 2 b c + c^2. So, 2 a c + 2 b c = -2 a c - 2 b c. Using b c = 0, 4 a c = 0 a c = 0. Hence a and c are perpendicular, not parallel. Therefore, statement (B) is incorrect. Now consider statement (A). Since a c = 0, | a + c |^2 = | a |^2 + 2 a c + ^2 | c |^2 = | a |^2 + ^2 | c |^2 | a |^2. Therefore, | a + c | | a | for all real. So statement (A) is correct. Therefore, the correct option is C. Given: a c = 0 from the first condition. Find: Whether | a + c | | a | for all real. Because a is perpendicular to c, | a + c |^2 = ( a + c) ( a + c). Expanding, | a + c |^2 = | a |^2 + 2 a c + ^2 | c |^2. Since a c = 0, | a + c |^2 = | a |^2 + ^2 | c |^2. Now ^2 | c |^2 0, so | a + c |^2 | a |^2. Taking square roots gives | a + c | | a |. Thus statement (A) is true, and statement (B) is false.

Let a = 2 i + j + k, and b and c be two nonzero vectors such that… — Solution | JEE Mathematics

MCQMediumJEE 2023Dot Product

Let $\mathbf{a} = 2\hat{i} + \hat{j} + \hat{k}$ , and $\mathbf{b}$ and $\mathbf{c}$ be two nonzero vectors such that

\left| \mathbf{a} + \mathbf{b} + \mathbf{c} \right| = \left| \mathbf{a} + \mathbf{b} - \mathbf{c} \right| \quad \text{and} \quad \mathbf{b} \cdot \mathbf{c} = 0.

Consider the following two statements: (A) The magnitude of the vector $\mathbf{a}$ plus $\lambda$ times the vector $\mathbf{c}$ is greater than or equal to the magnitude of vector $\mathbf{a}$ , for all real numbers $\lambda$ . (B) $\mathbf{a}$ and $\mathbf{c}$ are always parallel.

A
only (B) is correct
B
neither (A) nor (B) is correct
C
only (A) is correct
D
both (A) and (B) are correct

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $\mathbf{a} = 2\hat{i} + \hat{j} + \hat{k}$ , $\left| \mathbf{a} + \mathbf{b} + \mathbf{c} \right| = \left| \mathbf{a} + \mathbf{b} - \mathbf{c} \right|$ , and $\mathbf{b} \cdot \mathbf{c} = 0$ .

Find: Which of the statements (A) and (B) is correct.

Start with

\left| \mathbf{a} + \mathbf{b} + \mathbf{c} \right| = \left| \mathbf{a} + \mathbf{b} - \mathbf{c} \right|.

Squaring both sides,

\left( \mathbf{a} + \mathbf{b} + \mathbf{c} \right)^2 = \left( \mathbf{a} + \mathbf{b} - \mathbf{c} \right)^2.

Expanding,

\mathbf{a}^2 + 2\mathbf{a} \cdot \mathbf{b} + 2\mathbf{a} \cdot \mathbf{c} + \mathbf{b}^2 + 2\mathbf{b} \cdot \mathbf{c} + \mathbf{c}^2 = \mathbf{a}^2 + 2\mathbf{a} \cdot \mathbf{b} - 2\mathbf{a} \cdot \mathbf{c} + \mathbf{b}^2 - 2\mathbf{b} \cdot \mathbf{c} + \mathbf{c}^2.

So,

2 \mathbf{a} \cdot \mathbf{c} + 2 \mathbf{b} \cdot \mathbf{c} = -2 \mathbf{a} \cdot \mathbf{c} - 2 \mathbf{b} \cdot \mathbf{c}.

Using $\mathbf{b} \cdot \mathbf{c} = 0$ ,

4 \mathbf{a} \cdot \mathbf{c} = 0 \quad \Rightarrow \quad \mathbf{a} \cdot \mathbf{c} = 0.

Hence $\mathbf{a}$ and $\mathbf{c}$ are perpendicular, not parallel. Therefore, statement (B) is incorrect.

Now consider statement (A). Since $\mathbf{a} \cdot \mathbf{c} = 0$ ,

\left| \mathbf{a} + \lambda \mathbf{c} \right|^2 = \left| \mathbf{a} \right|^2 + 2\lambda \, \mathbf{a} \cdot \mathbf{c} + \lambda^2 \left| \mathbf{c} \right|^2 = \left| \mathbf{a} \right|^2 + \lambda^2 \left| \mathbf{c} \right|^2 \ge \left| \mathbf{a} \right|^2.

Therefore,

\left| \mathbf{a} + \lambda \mathbf{c} \right| \ge \left| \mathbf{a} \right|

for all real $\lambda$ . So statement (A) is correct.

Therefore, the correct option is C.

Why Statement (A) Holds

Given: $\mathbf{a} \cdot \mathbf{c} = 0$ from the first condition.

Find: Whether $\left| \mathbf{a} + \lambda \mathbf{c} \right| \ge \left| \mathbf{a} \right|$ for all real $\lambda$ .

Because $\mathbf{a}$ is perpendicular to $\mathbf{c}$ ,

\left| \mathbf{a} + \lambda \mathbf{c} \right|^2 = \left( \mathbf{a} + \lambda \mathbf{c} \right) \cdot \left( \mathbf{a} + \lambda \mathbf{c} \right).

Expanding,

\left| \mathbf{a} + \lambda \mathbf{c} \right|^2 = \left| \mathbf{a} \right|^2 + 2\lambda \, \mathbf{a} \cdot \mathbf{c} + \lambda^2 \left| \mathbf{c} \right|^2.

Since $\mathbf{a} \cdot \mathbf{c} = 0$ ,

\left| \mathbf{a} + \lambda \mathbf{c} \right|^2 = \left| \mathbf{a} \right|^2 + \lambda^2 \left| \mathbf{c} \right|^2.

Now $\lambda^2 \left| \mathbf{c} \right|^2 \ge 0$ , so

\left| \mathbf{a} + \lambda \mathbf{c} \right|^2 \ge \left| \mathbf{a} \right|^2.

Taking square roots gives

\left| \mathbf{a} + \lambda \mathbf{c} \right| \ge \left| \mathbf{a} \right|.

Thus statement (A) is true, and statement (B) is false.

Common mistakes

Assuming that equal magnitudes directly imply the vectors inside are equal or parallel. This is wrong because only the lengths are equal. Square the magnitudes and expand the dot products instead.
Concluding that $\mathbf{a}$ and $\mathbf{c}$ are parallel from $\mathbf{a} \cdot \mathbf{c} = 0$ . This is wrong because zero dot product means the vectors are perpendicular. Use the dot product interpretation correctly.
Accepting statement (A) without using the derived condition $\mathbf{a} \cdot \mathbf{c} = 0$ . The inequality is not true for arbitrary vectors, but it is true here because the cross term vanishes. First prove orthogonality, then evaluate the magnitude.

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Answer

Step-by-step solution

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