The sum and difference of two vectors A and B are A - B -2- -6- -k- - A - B -4- -2- -11k- Find the magnitude of each vector and their scalar product A - B

A⃗ + B⃗ = #160; 2î + 6ĵ + k̂ #160; #160; #160; #160; #160;...(i) A⃗ B⃗ = #160; 4î + 2ĵ 11k̂ #160; #160; #160; ...

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Byju's Answer

Standard XII

Physics

Vector Addition

The sum and d...

Question

The sum and difference of two vectors $\to A$ and $\to B$ are $\to A + \to B = 2^i + 6^j +^k; \to A - \to B = 4^i + 2^j - 11^k$ . Find the magnitude of each vector and their scalar product $\to A . \to B$

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Solution

$\to A + \to B$ = $2^i + 6^j +^k$ ...(i)
$\to A - \to B$ = $4^i + 2^j - 11^k$ ...(ii)
Adding equations (i) and (ii) :
$2 \to A$ = $6^i + 8^j - 10^k$
$\to A$ = $3^i + 4^j - 5^k$
Subtracting (ii) from (i)
$2 \to B$ = $- 2^i + 4^j + 12^k$
$\to B$ = $-^i + 2^j + 6^k$
Therefore, $∣ ∣ \to A ∣ ∣$ = $5 \sqrt{2}$
$∣ ∣ \to B ∣ ∣$ = $\sqrt{41}$
$\to A . \to B$ = $- 3 + 8 - 30$
= $- 25$

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\to A + \to B = 2^i + 6^j + 11^k

and

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The scalar product of the vector $\to a =^i +^j +^k$ with a unit vector along the sum of vector $\to b = 2^i + 4^j + 5^k$ and $\to c = λ^i + 2^j + 3^k$ is equal to one. Find the value of $λ$ and hence find the unit vector along $\to b + \to c$ .

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The sum and difference of two vectors →A and →B are →A+→B=2^i+6^j+^k;→A−→B=4^i+2^j−11^k. Find the magnitude of each vector and their scalar product →A.→B

→A+→B = 2^i+6^j+^k ...(i)→A−→B = 4^i+2^j−11^k ...(ii)Adding equations (i) and (ii) :2→A = 6^i+8^j−10^k→A = 3^i+4^j−5^kSubtracting (ii) from (i)2→B = −2^i+4^j+12^k→B = −^i+2^j+6^kTherefore, ∣∣→A∣∣ = 5√2∣∣→B∣∣ = √41→A.→B = −3+8−30 = −25

The sum and difference of two vectors $\to A$ and $\to B$ are $\to A + \to B = 2^i + 6^j +^k; \to A - \to B = 4^i + 2^j - 11^k$ . Find the magnitude of each vector and their scalar product $\to A . \to B$