If either a → = 0 → or b → = 0 → , then a → × b → = 0 → . Is the converse true? Justify your answer with an example.

Converse of the statement is If a → × b → = 0 , then a → = 0 or b → = 0 . Let us check whether it is true or not. Let a → = 2 i ^ + 3 j ^ + 4 k ^ , b → = 4 i ^ + 6 j ^ + 8 k ^ To show: a → × b → = 0 → For a → × b → = i ^ j ^ k ^ 2 3 4 4 6 8 = i ^ 3 × 8 - 4 × 6 - j ^ 2 × 8 - 4 × 4 + k ^ 2 × 6 - 3 × 4 = i ^ ( 24 - 24 ) - j ^ ( 16 - 16 ) + k ^ ( 12 - 12 ) = 0 i ^ + 0 j ^ + 0 k ^ | a → | = 2 2 + 3 2 + 4 2 = 29 ∴ a → ≠ 0 → | b → | = 4 2 + 6 2 + 8 2 = 116 ∴ b → ≠ 0 → Hence, the converse of the statement need not to be true.

If a → = a 1 i ^ + a 2 j ^ + a 3 k ^ , b → = b 1 i ^ + b 2 j ^ + b 3 k ^ and c → = c 1 i ^ + c 2 j ^ + c 3 k ^ , then verify that a → × ( b → + c → ) = a → × b → + a → × c →

Given a → = a 1 i ^ + a 2 j ^ + a 3 k ^ , b → = b 1 i ^ + b 2 j ^ + b 3 k ^ and c → = c 1 i ^ + c 2 j ^ + c 3 k ^ . Consider LHS, i.e., a → × b → + c → = ( a 1 i ^ + a 2 j ^ + a 3 k ^ ) × [ ( b 1 i ^ + b 2 j ^ + b 3 k ^ ) + ( c 1 i ^ + c 2 j ^ + c 3 k ^ ) ] ⇒ a → × b → + c → = ( a 1 i ^ + a 2 j ^ + a 3 k ^ ) × ( ( b 1 + c 1 ) i ^ + ( b 2 + c 2 ) j ^ + ( b 3 + c 3 ) k ^ ) ⇒ a → × b → + c → = i ^ j ^ k ^ a 1 a 2 a 3 ( b 1 + c 1 ) ( b 2 + c 2 ) ( b 3 + c 3 ) ⇒ a → × b → + c → = a 2 ( b 3 + c 3 ) - a 3 ( b 2 + c 2 ) i ^ - a 1 ( b 3 + c 3 ) - a 3 ( b 1 + c 1 ) j ^ + a 1 ( b 2 + c 2 ) - a 2 ( b 1 + c 1 ) k ^ ⇒ a → × b → + c → = ( a 2 b 3 + a 2 c 3 - a 3 b 2 - a 3 c 2 ) i ^ - ( a 1 b 3 + a 1 c 3 - a 3 b 1 - a 3 c 1 ) j ^ + ( a 1 b 2 + a 1 c 2 - a 2 b 1 - a 2 c 1 ) k ^ . . . i Now consider RHS, i.e., a → × b → + a → × c → = ( a 1 i ^ + a 2 j ^ + a 3 k ^ ) × ( b 1 i ^ + b 2 j ^ + b 3 k ^ ) + ( a 1 i ^ + a 2 j ^ + a 3 k ^ ) × ( c 1 i ^ + c 2 j ^ + c 3 k ^ ) ⇒ a → × b → + a → × c → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 + i ^ j ^ k ^ a 1 a 2 a 3 c 1 c 2 c 3 ⇒ a → × b → + a → × c → = ( a 2 b 3 - a 3 b 2 ) i ^ - ( a 1 b 3 - a 3 b 1 ) j ^ + ( a 1 b 2 - a 2 b 1 ) k ^ + ( a 2 c 3 - a 3 c 2 ) i ^ - ( a 1 c 3 - a 3 c 1 ) j ^ + ( a 1 c 2 - a 2 c 1 ) k ^ ⇒ a → × b → + a → × c → = ( a 2 b 3 + a 2 c 3 - a 3 b 2 - a 3 c 2 ) i ^ - ( a 1 b 3 + a 1 c 3 - a 3 b 1 - a 3 c 1 ) j ^ + ( a 1 b 2 + a 1 c 2 - a 2 b 1 - a 2 c 1 ) k ^ . . . ii From the Equations i & ii LHS = RHS Therefore, a → × b → + c → = a → × b → + a → × c → Hence prove.

Find all vectors of magnitude 10 3 that are perpendicular to the plane of i ^ + 2 j ^ + k ^ and - i ^ + 3 j ^ + 4 k ^ .

Let a → = i ^ + 2 j ^ + k ^ and b → = - i ^ + 3 j ^ + 4 k ^ Unit vectors perpendicular to both a → and b → = ± a → × b → | a → × b → | Now, a → × b → = i ^ j ^ k ^ 1 2 1 - 1 3 4 = 5 i ^ - 5 j ^ + 5 k ^ ∴ | a → × b → | = | 5 i ^ - 5 j ^ + 5 k ^ | = 5 2 + ( - 5 ) 2 + 5 2 = 75 = 5 3 Unit vectors perpendicular to both a → and b → = ± 5 i ^ - 5 j ^ + 5 k ^ 5 3 = ± i ^ - j ^ + k ^ 3 Hence, the required vectors = 10 3 ± i ^ - j ^ + k ^ 3 = ± 10 ( i ^ - j ^ + k ^ ) .

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MATHEMATICS CLASS XII VOLUME-2>Chapter 6 - Vector or Cross Product>EXERCISE>Q 31

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The two adjacent sides of a parallelogram are $2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $\hat{i} - 2 \hat{j} - 3 \hat{k} .$ Find the unit vector parallel to one of its diagonals. If its area is $11 \sqrt{k}$ square units. Then the value of $k$ is

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Important Questions on Vector or Cross Product

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MATHEMATICS CLASS XII VOLUME-2>Chapter 6 - Vector or Cross Product>EXERCISE>Q 32

If either

\vec{a} = \vec{0}

or

\vec{b} = \vec{0},

then

\vec{a} \times \vec{b} = \vec{0} .

Is the converse true? Justify your answer with an example.

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MATHEMATICS CLASS XII VOLUME-2>Chapter 6 - Vector or Cross Product>EXERCISE>Q 33

If

\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}

and

\vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k},

then verify that

\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}

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MATHEMATICS CLASS XII VOLUME-2>Chapter 6 - Vector or Cross Product>EXERCISE>Q 34

If the area of the triangle with vertices: $A (1, 1, 2), B (2, 3, 5)$ and $C (1, 5, 5)$ is $\frac{\sqrt{k}}{2}$ square units. Find the value of $k$ .

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MATHEMATICS CLASS XII VOLUME-2>Chapter 6 - Vector or Cross Product>EXERCISE>Q 35

The area of the triangle with vertices: $A (1, 2, 3), B (2, - 1, 4)$ and $C (4, 5, - 1)$ is $\frac{\sqrt{k}}{2}$ . Find the value of $k$ .

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MATHEMATICS CLASS XII VOLUME-2>Chapter 6 - Vector or Cross Product>EXERCISE>Q 35

Find all vectors of magnitude $10 \sqrt{3}$ that are perpendicular to the plane of $\hat{i} + 2 \hat{j} + \hat{k}$ and $- \hat{i} + 3 \hat{j} + 4 \hat{k}$ .