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Vector Algebra question

2023 · 11 Apr · Shift 1 · Q32
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  5. /2023 · 11 Apr · Shift 1 · Q32

Vector Algebra question

2023 · 11 Apr · Shift 1 · Q32

JEE MainMathematicsVector AlgebraMCQ+4 / −1
For any vector a⃗=a1i^+a2j^+a3k^\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}a=a1​i^+a2​j^​+a3​k^, with 10∣ai∣(A):10\left|a_{i}\right|(A):10∣ai​∣(A):\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\} \leq|\vec{a}|(B):(B) :(B):|\vec{a}| \leq 3 \max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\}$$
  1. A
    Only (B) is true
  2. B
    Only (A) is true
  3. C
    Neither (A) nor (B) is true
  4. D
    Both (A) and (B) are true
View written solutionFree

Correct answer: D

  1. Let
a⃗=a1i^+a2j^+a3k^\vec a=a_1\hat i+a_2\hat j+a_3\hat ka=a1​i^+a2​j^​+a3​k^

Then its magnitude is

∣a⃗∣=a12+a22+a32.|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}.∣a∣=a12​+a22​+a32​​.

Let

M=max⁡{∣a1∣,∣a2∣,∣a3∣}.M=\max\{|a_1|,|a_2|,|a_3|\}.M=max{∣a1​∣,∣a2​∣,∣a3​∣}.

We check statements (A) and (B).

  1. Checking (A):

Since MMM is the maximum of ∣a1∣,∣a2∣,∣a3∣|a_1|,|a_2|,|a_3|∣a1​∣,∣a2​∣,∣a3​∣, one of the terms ∣ai∣|a_i|∣ai​∣ equals MMM. Therefore,

a12+a22+a32≥M2.a_1^2+a_2^2+a_3^2 \ge M^2.a12​+a22​+a32​≥M2.

Taking square root on both sides,

∣a⃗∣=a12+a22+a32≥M.|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}\ge M.∣a∣=a12​+a22​+a32​​≥M.

Hence,

max⁡{∣a1∣,∣a2∣,∣a3∣}≤∣a⃗∣.\max\{|a_1|,|a_2|,|a_3|\}\le |\vec a|.max{∣a1​∣,∣a2​∣,∣a3​∣}≤∣a∣.

So (A) is true.

  1. Checking (B):

Since each of ∣a1∣,∣a2∣,∣a3∣≤M|a_1|,|a_2|,|a_3|\le M∣a1​∣,∣a2​∣,∣a3​∣≤M, we have

a12≤M2,a22≤M2,a32≤M2.a_1^2\le M^2,\quad a_2^2\le M^2,\quad a_3^2\le M^2.a12​≤M2,a22​≤M2,a32​≤M2.

Adding,

a12+a22+a32≤3M2.a_1^2+a_2^2+a_3^2\le 3M^2.a12​+a22​+a32​≤3M2.

Taking square root,

∣a⃗∣=a12+a22+a32≤3 M.|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}\le \sqrt{3}\,M.∣a∣=a12​+a22​+a32​​≤3​M.

Now since

3≤3,\sqrt{3}\le 3,3​≤3,

it follows that

∣a⃗∣≤3M.|\vec a|\le 3M.∣a∣≤3M.

Hence,

∣a⃗∣≤3max⁡{∣a1∣,∣a2∣,∣a3∣}.|\vec a|\le 3\max\{|a_1|,|a_2|,|a_3|\}.∣a∣≤3max{∣a1​∣,∣a2​∣,∣a3​∣}.

So (B) is also true.

  1. Conclusion:

Both (A) and (B) are true.

Therefore the correct option is

D\boxed{D}D​
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